Replace string in multiple files (same directory) with condition

This should do the job :

sed -i 's/^Code: 0$/Code: 1/' $(grep -l '^Number: [23]' File*)

• the sed part does the actual replacements
• the grep -l ... File* part lists all files with name matching 'File*' and having a line starting with either Number: 2 or Number: 3
• the $(...) construct surrounding the grep command is a command substitution that will feed files listed by grep into sed

With GNU tools:

grep -lxZ 'Number: [23]..' ./File* |
  xargs -r0 sed -i 's/^Code: 0$/Code: 1/'

Assuming Number always precedes Code, and Code might have any value, then using sed:

sed -i '/^Number: [23]/{n;s/^Code: .*$/Code 1/}' File*

If the content is consistent, so that the line after Number is always Code 0, that can be simplified to:

sed -i '/^Number: [23]/{n;s/0/1/}' File*

Do not use sed for this task without creating temporary files because redirecting back to the source would destroy the file's contents. Instead, use the ed editor.
I assume that filenames are in the range [a-zA-Z0-9_-]

% cat scr.ed
1,$s/Code: 0/Code: 1/
w
q
% cat scr.sh
#!/bin/sh
for file in `grep -l '^Number: [23]' File*`
do
    ed -s $file <scr.ed
done

% chmod 755 scr.sh
% ./scr.sh