That seems a bit over complicated.

print $_, m[\s*(\d{0,3}\b)?] 
    ? " contained '$1'" 
    : ' contained nothing that matched' 
    for '','1','1 ',' 1','12',' 12 ','123',' 123',' 123 ','1234','1 2 
+3';

Use of uninitialized value in concatenation (.) or string at (eval 21)
+ line 1, <STDIN> line 19.
  contained ''
1  contained '1'
1   contained '1'
 1  contained '1'
12  contained '12'
 12   contained '12'
123  contained '123'
 123  contained '123'
 123   contained '123'
1234  contained ''
1 2 3  contained '1'
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---

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"Efficiency is intelligent laziness." -David Dunham
"Think for yourself!" - Abigail
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That doesn't match "12 ", which you said should be matched.

Also, \s should be used instead of \W (unless you want to match things like "!@#").

hum, it does matches all of those

print(   "1  " =~ /\d{0,3}/   ?"match":"no match", $/);
print(   "12 " =~ /\d{0,3}/   ?"match":"no match", $/);
print(   "  1" =~ /\d{0,3}/   ?"match":"no match", $/);
print(   "   " =~ /\d{0,3}/   ?"match":"no match", $/);
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Now, if you want the regexp to completely match exactly three characters, consiting of 0+ digits padded with spaces on either side, that's much more complicated.

We can write it out the long way:
/^(?:   |  \d| \d\d|\d\d\d|\d  |\d\d |\d  )$/
This doesn't compress well.

/^[ \d]{3}$/
will match them all, but it will also match "1 2".

/^(?: [ \d]{2}|[ \d]{2} |\d{3})$/ will work.

Anything else I can think of invovles potentially matching more than three characters (/^\s*\d{0,3}\s*$/) or involves more than a regexp.

If the spaces can only be leading spaces, then the long form would be:
/^(?:   |  \d| \d\d|\d\d\d)$/
which doesn't really simplify.

btw, you should be using \D if you mean non-digit, or \S if you mean non-whitespace. \W doesn't appear to be appropriate here.