Pyth, 7 bytes

eSm.OCM

Test suite

Explanation:

eSm.OCM   | Full program
eSm.OCMdQ | with implicit variables
----------+-------------------------------
  m     Q | For each word in the input,
     CMd  |  Get the ord of each character
   .O     |  Get the average
eS        | Get the max

Python, 46 bytes

lambda l:max(sum(map(ord,s))/len(s)for s in l)

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Unfortunately, the only built-in mean function Python has is in the statistics module, and that's too expensive to import.

K (ngn/k), 18 11 bytes

|/{+/x%#x}'

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Massive -7 bytes golf thanks to Steffan!

Damn everyone is answering quick. Anyways here's my solution. Takes input as a list of strings.

Explanation:

|/{+/x%#x}'  Main function. Takes implicit input
          '  For each string in the input...
  {      }   Execute a function that...
   +/x       Takes sum of every character in the string
      %      And divide it by
       #x    The length of the string
|/           Get the maximum value

Wren, 68 bytes

Fn.new{|x|x.reduce(0){|y,z|y.max(z.bytes.reduce{|a,b|a+b}/z.count)}}

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APL(Dyalog Unicode), 14 13 bytes ^SBCS

-1 bytes thanks to Adám

⌈/+⌿∘⎕UCS¨÷≢¨

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A tacit function which finds the maximum ord. Originally wrote a dfn, but I tried random stuff to convert it to tacit and it somehow worked out.

⌈/+⌿∘⎕UCS¨÷≢¨
          ¨              ⍝  for each element in the input...
     ⎕UCS               ⍝  convert the characters to their ascii values...
  +⌿∘                   ⍝  and sum the resulting ascii values...
           ÷             ⍝  divided by...
            ≢¨           ⍝  the length of each element...
⌈/                       ⍝  and take the maximum of that

Jelly, 4 bytes

OÆmṀ

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><>, 80 51 48 bytes

0>0001.
$i:"!"(?v+$1+
(@:{:,$&/
v?(0&~$?/.02
\n;

Animated Version

Requires a extra space at the end of the input. Also assumes no 2 spaces in a row. (no 0 length words). Anything with ORD of 32 or lower is considered white-space.

Explanation

Top row: Push 3 0s on to the stack. The first is the current best ORD, the second is the length of the current word, and the last is the sum of the ord values of the current word. 01. is a jump to the next row.

The second row sums a word. If a character is less than "!" (char value 33) go down. Otherwise sum the word and increment the length. `$1+

In the third row, we fist push the last value to the register & . We use this to later check if this was the end of the string. Then we divide by the length to get the new ORD value. Then we use the :{:@ trick to copy the stack if it has only 2 elements (which is always the case) from the stack and compare them. If the new one is better we swap the stack with $. In either case we delete the top item.

Lastly, we take the value of the register and check if it is negative, &0(?. If so, we print the maximum ord and exit. If not, we jump back to 00. To get the correct direction we use the 00!.| trick.

><>, 65 45 44 40 bytes

000i:"!"(?v+$1+$20.
$~$0(?n00.>@$,:{:@(?

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Takes input as a space separated string.

Factor, 24 bytes

[ 0 [ mean max ] foldl ]

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Just fold with the max of the mean.

R, 38 bytes

\(x)max(sapply(Map(utf8ToInt,x),mean))

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               Map(utf8ToInt,x)        # for every string in the input, store the vector of its codepoints in a list
        sapply(                ,mean)  # apply mean to each of the list elements and return a vector
\(x)max(                             ) # take max

05AB1E, 10 9 4 bytes

ÇÅAà

-5 thanks to Sʨɠɠan

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Explained:

ÇÅAà  # Implicit input as a list
      # Implicit map over the input:
Ç     #   Get the ord of each character
 ÅA   #   Get the average of the ords
   à  # Get the maximum value

Pip, 12 bytes

M:$+A*_/#_Mg

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Knight (v2), 56 bytes

;=mF;W=pP;=s!=i~1;W>Lp=i+1i=s+*10A Gp iTs&<m=t/sLp=m tOm

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Because Knight doesn't support floating point values, it outputs the maximum average ord multiplied by \$10\$, as allowed by the rules:

It may be rounded to any number (\$\ge1\$) of decimal places. If your language does not support floating points, you may output the average multiplied by 10.

Julia, 34 bytes

!a=max(sum.(Int,a)./length.(a)...)

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Excel (ms365), 70 bytes

Formula in F1:

=MAX(MAP(A1:C1,LAMBDA(a,SUM(CODE(MID(a,SEQUENCE(LEN(a)),1)))/LEN(a))))

MathGolf, 4 bytes

$m▓╙

Input as a list of lists of characters.

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Explanation:

$     # Get the codepoint of each inner-most character of the (implicit) input-list
 m    # Map over each list of integers:
  ▓   #  Pop and push the average of the list
   ╙  # After the map: pop and push the maximum
      # (after which the entire stack is output implicitly as result)

J, 18 bytes

[:>./(1#.3&u:%#)&>

Accepts list of boxed strings

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[:>./(1#.3&u:%#)&>
[:                  NB. cap, [: f g y -> f (g y)
                &>  NB. for each input item, result will be unboxed
     (         )    NB. monadic fork
              #     NB. length
         3&u:       NB. convert input to list of char codes
             %      NB. vectorized division
      1#.           NB. sum the result to compute average
  >./               NB. max item

PHP 8.x, 75 73 bytes

Creates an anonymous function that returns the expect values, hopefully.

fn($x)=>max(array_map(fn($z)=>array_sum(unpack('C*',$z))/strlen($z),$x));

-2 bytes thanks to Sʨɠɠan.

How does it work?

This uses the unpack('C*', ...) function to convert all characters into an unsigned char (0-255).
This assumes all characters are part of the ASCII table.

Then it calculates the average using array_sum([...])/strlen([...]).
The function array_sum() takes an array and returns the sum of all elements, and strlen() returns the length of the string.

The function max() takes the array with the averages and returns the highest value from it.

Anonymous functions implicitly return values.

Example usage

You need to assign this to a variable, or call with call_user_func() or call_user_func_array():

<?php
$fn = fn($x)=>max(array_map(fn($z)=>array_sum(unpack('C*',$z))/strlen($z),$x));

// Should output: float(110.4)
var_dump($fn(['hello', 'world', 'bye']));

You can try this here (with test examples): https://onlinephp.io/c/2a424

C (clang), ^{87 86 84} 81 bytes.

v;*p;f(**t,float*r){for(*r=0;p=*t;*r=fmax(*r,v*1./(p-*t++)))for(v=0;*p;)v+=*p++;}

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MATLAB, 39 31 bytes

@(s)max(cellfun(@(x)mean(x),s))

Matlab unfortunately doesn't convert strings ("") directly to ASCII values using the double function, and if converted to char type then a string array will be padded with spaces to make a proper matrix. As a result, the input is a cellstr type, which needs to be looped over using the cellfun function with an anonymous function that takes the average of the unicode values of the strings (''). The output is then the maximum of those values.

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edit: -8 bytes thanks to Guiseppe

T-SQL, 125 bytes

WITH C as(SELECT len(x)m,0b,*FROM @
UNION ALL SELECT m-1,b+ascii(right(x,m)),x
FROM C WHERE
m>0)SELECT max(b*1./len(x))FROM c

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Raku, 27 bytes

*».&{.ords.sum/.chars}.max

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Edit: as per advice from @Steffan

Husk, 6 bytes

▲moAmc

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Outputs the highest average ord value as an exact fraction.

▲moAmc
 mo    # map 2 functions over each element of input
    mc #   get ord values of each character
   A   #   get the average of those
▲      # output the maximum.

JavaScript (Node.js), 58 bytes

a=>Math.max(...a.map(s=>eval(Buffer(s).join`+`)/s.length))

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Ruby, 48 bytes

->l{l.map{_1.chars.sum(&:ord)/_1.size.to_f}.max}

As the division of two integers in ruby gives an integer, and what was expected was a float, it's long :/
And .fdiv is one byte longer

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Sequences, \$9 \log_{256}(96) \approx 7.41\$ bytes

[$v$aH]gM

Explanation

[$v$aH]gM  // Implicit list of string input
[     ]    // Loop through the input list:
 $v$       //   Get a list of ords of each character
    aH     //   Get the average and append to `g`
       g   // Push the list `g`
        M  // Get the maximum value
           // Implicit output

sclin, 38 bytes

"dup S>c0\+ fold rev len /"map0\| fold

Try it here! This might be a good case for adding sum/prod/min/max functions to sclin, the fold construct is alright but it could be better.

For testing purposes:

["hello" "world" "bye"] ; 30N>d n>o
"dup S>c0\+ fold rev len /"map0\| fold

Explanation

Prettified code:

( dup S>c 0 \+ fold rev len / ) map 0 \| fold

Assuming input list xs.

• (...) map map over xs...
  • dup S>c get codepoints
  • 0 \+ fold sum
  • rev len / divide by length (i.e. average)
• 0 \| fold maximum

Mathematica, 29 bytes

Max[Mean/@ToCharacterCode@#]&

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Fig, \$6\log_{256}(96)\approx\$ 4.939 bytes

]KemMC

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]KemMC
     C - Charcodes
  emM  - Mean of each
 K     - Sort
]      - Last