Answer     7

We note that the ending digit of $3^k$ repeats every $4$ terms: $3$, $9$, $7$, $1$, $3$, $\cdots$. Because $2019\equiv 3\pmod{4}$, thus the ending digits of $13^{2019}$ is the same as $3^3$ which is $\boxed{7}$.

Alternatively, it is also possible to solve this problem using the $(-1)$ technique. $$13^{2019}\equiv 3^{2019}\equiv 9^{1009}\times 3 \equiv (-1)^{1009}\times 3 \equiv \boxed{7}\pmod{10}$$