Um, $ 2 \equiv -1 \pmod {3}$. $ 2^a \equiv \pm 1 \pmod {3}$. So $ 2^a$ does not divide $ 3$. Regardless of permuting the digits, the number will still not be divisible by 3 because the divisibility rule relies only on the sum of the digits. So the number will not be divisible by 3, much less be a power of 3. Am I reading the question wrong? The problem seems too easy.

Am I reading the question wrong? The problem seems too easy.