By Fermat's Little Theorem, we have $a^{28}\equiv 1 \pmod{29}$ for all positive integers $a$ which are not multiples of $29$.

It follows that $a^{14}\equiv\pm 1 \pmod{29}$, so $a^{15}\equiv\pm a \pmod{29}$ for all such $a$. Therefore, if $a^{15}\equiv 20 \pmod{29}$, then $\pm a\equiv 20 \pmod{29}$.

The first few positive integers that satisfy $\pm a\equiv 20\pmod{29}$ are $9$, $20$, $38$, $\cdots$.

We know that $9^{14} = 3^{28}\equiv 1 \pmod{29}$, so $9^{15}\equiv 9 \pmod{29}$.

Next we try $a = 20$, and we find that $20^{14}\equiv (20+29)^{14}\equiv 7^{28}\equiv 1 \pmod{29}$, and so $20^{15}\equiv 20 \pmod{29}$. Therefore, the answer is $\boxed{20}$.