Regular expression - every b preceded and followed by an even number of a's
The answer you already have covers almost all the cases: any time there's a $b$, there need to be an even number of $a$'s, so that covers all words containing a $b$.

However, there's an edge case where there are no $b$'s in your word, so you need to union your language with the case where the word has no $b$'s.

$((aa)^*b(aa)^*)^* \cup a^*$ should fit your final answer.