Trivially $10 \mid a$, and as $10$ and $30$ don't work, and everything over $40$ works, thus the answer is all multiples of $10$ except for $10,30$.

Since $20^{19}=2^{38}5^{19}$, such an integer $a$ must be divisible by $2$ and by $5$. * $a=10$ does not work, as $10^{10}$ is not divisible by $2^{38}$. * $a=20$ does work, as $20^{20}$ is divisible by $20^{19}$. * $a=30$ does not work, as $30^{30}$ is divisible by $2^{30}$ but not by $2^{38}$. * Finally, every $a=10k$ with $k\ge4$ works, as $a^a$ is then divisible by $2^{40}$ and by $5^{40}$.