If it was $aba^{-1}$ = $b^2$ I could use the theorem that if $aba^{-1}$ = $b^n$ then $a^kba^{-k}$ = $b^{n^k}$ so that $b$ = $ebe$ = $a^2ba^{-2}$ = $b^4$ so that $|b|$ = 3. Is there a similar method here?