Number Theory 20: Since $a = 2^n \cdot 3^m$, $a^6 = 2^{6n} \cdot 3^{6m}$, and $$\begin{aligned} 6^a &= (2 \cdot 3)^a \\ &= (2 \cdot 3)^{2^n \cdot 3^m} \\ &= 2^{2^n \cdot 3^m} \cdot 3^{2^n \cdot 3^m} \end{aligned}.$$ Look at the exponents of both. What do you need to ask of $n$ and $m$ so that $a^6 \mid 6^a$?