Pumping lemma doesn’t provide us with an if and only if condition for regularity. So, if a language doesn’t satisfy pumping lemma then it isn’t regular, but converse is not true.
So, I don’t think we can use pumping lemma to show that a language is regular.
To show given language is regular, you can use Myhill-Nerode theorem, simply come up with a DFA/NFA or a regular expression, or use some closure properties.
For the given language, you can show its regularity using closure property of regular languages (intersection, to be precise):
$0^*1^*$ is regular, and so is the $L_{\geq 20} = \{w : |w| \geq 20\}$. It’s quite easy to see that the given language is intersection of these two languages, and hence it is regular.