jmc

Notice that for each integer $n$ satisfying $1\le n\le 19,n\ne 10$, then using the sum of cubes factorization, $n^3+(20-n{)}^3=(n+20-n)(n^2+(20-n)n+(20-n{)}^2)$ $=20(n^2+(20-n)n+(20-n{)}^2)$. Thus, we can pair each integer with the difference between $20$ and that integer, and the sum of their cubes is divisible by $20$. Hence, the units digit of the sum of the first $19$ cubes, excluding $10$, is equal to $0$. Also, the units digits of $10^3$ and $20^3$ are clearly $0$, so we just need to find the units digit of the cube of $21$, which is $\boxed{1}$.