Let $\overline{abc}$ be such a number. The condition is $\overline{abc} = 13 \cdot \overline{ac}$, that is $100a + 10b + c = 130a + 13c$, same as $10b = 30a + 12c$. We notice that $5$ divides both $10b$ and $30a$, so $12c$ is divisible by $5$, hence $c \in \{0, 5\}$.

If $c = 0$, then $b = 3a$. Since $a$ is a non-null digit, $(a = 1, b = 3)$, $(a = 2, b = 6)$ and $(a = 3, b = 9)$, which lead to the solutions $130$, $260$ and $390$.

If $c = 5$, then $b = 3a + 6$. Since $a$ is a non-null digit, the only possibility is $(a = 1, b = 9)$, which leads to the solution $195$.