Solution:
Let $\underline{x}_{k}$ denote the number $x$ in base $k$. Observe that if $f\left(\underline{x}_{3}\right)=\underline{y}_{3}$, then
$$f\left(\underline{x 0}_{3}\right)=f(3 x)=3 f(x)=\underline{y 0}_{3}$$
and
$$f\left(\underline{x 1}_{3}\right)=f\left(\underline{x 2}_{3}\right)=3 f(x)+1=\underline{y 1}_{3} .$$
Thus, $\underline{f(n)}_{3}$ is simply $\underline{n}_{3}$ with all 2's replaced with 1's. We also see that $13=111_{3}$. Thus, $f(m)=13$ if and only if $m=\underline{a b c_{3}}$ for digits $a, b, c \in\{1,2\}$. Each of $a, b$, and $c$ takes on each possible value exactly 4 times, so the sum is
$$(4 \cdot 1+4 \cdot 2)\left(3^{2}+3^{1}+3^{0}\right)=156$$