Solution:

By repeatedly multiplying by $3$, we see that the units digits of powers of $3$ are either $3$, $9$, $7$, or $1$. Suppose that $N$ is a power of $3$ with an even tens digit; we will prove that the tens digit of $3N$ is also even, from which it will follow inductively that every power of $3$ from $27$ onward has an even tens digit.

If $N$ ends in $1$ or $3$, then when $N$ is tripled, there will be no carrying from the units place to the tens place. Then the tens digit of $3N$ will arise from tripling the tens digit of $N$ and hence will be even.

If $N$ ends in $7$ or $9$, then there will be a carry of $2$ from the units place to the tens place. The tens digit of $3N$ will arise from tripling the tens digit of $N$ and hence will still be even.