How can I extract a number from a command's output?

Assuming you mean to print the number between @<whitespaces> and GHz at the end of lines that start with model name, that would be:

sed -nE 's/^model name.*@[[:space:]]*([.[:digit:]]+)GHz$/\1/p' /proc/cpuinfo

Or with GNU grep:

grep -Po '^model name.*@\s*\K[.\d]+(?=GHz$)' /proc/cpuinfo

With lscpu, see also:

lscpu -J | jq -r '
  .lscpu[] |
  select(.field == "Model name:").data |
  match("@\\s*([\\d.]+)GHz$").captures[0].string |
  tonumber'

Or:

lscpu -J | jq -r '.lscpu[]|select(.field == "CPU max MHz:").data|tonumber'

For the max frequency in MHz instead of the base frequency from the model name.

Using sed (GNU sed for its I flag for case insensitive matching).

$ lscpu | sed -En 's/^model name.* ([0-9.]+).*/\1/Ip'
2.60

You can try something like this with awk:

awk '{print int($NF*100)/100.0}' /proc/cpuinfo

this line can be rewrited but I am not sure how different awk implementations will handle it:

awk '{print $NF*1.0}' /proc/cpuinfo

This will print 2.6. If you want to have it with zero you can use:

awk '{printf "%4.2f",int($NF*100)/100.0}' /proc/cpuinfo

If you want to use lscpu you can it on this way:

lscpu|awk '/odel name/ {print int($NF*100)/100.0}'

<command>| awk '{gsub(/[a-zA-Z]/,"",$NF);print $NF}'

Tested with echoing the following command output
echo "model name      : Intel(R) Xeon(R) CPU E5-2690 v4 @ 2.60GHz"| awk '{print $NF}'|sed "s/[a-zA-Z]//g"

output

2.60

LANG=C lscpu | sed -rn "s/.*Model name.* @ (.*)GHz/\1/p"

The LANG=C is needed, if your language setting isn't set to English.

The -n for sed suppresses output by default. The -r is for simpliefied typing of capturing groups (without backslash).

The "s/A/B/p" command substitutes A with B and the p performs a print, if condition A is met (since we suppressed output by default, that's neccessary).

A is the regex, and \1 the first (and only) capture group.