jmc

We start by computing the sum of the units digits of all multiples of $3$ between $0$ and $30$. Excluding $0$, every possible digit appears exactly once as a unit digit of a multiple of $3$: the set of multiples of $3$ between $0$ and $30$ consists of the numbers $0,3,6,9,12,15,18,21,24,27,30$. Thus, the sum of their units digits is equal to $$1+2+3+4+5+6+7+8+9=\frac{9\cdot 10}{2}=45.$$ We must sum the units digits of the multiples of $3$ between $31$ and $50$. The relevant multiples of $3$ are $33,36,39,42,45,48$, and the sum of their units digits is $3+6+9+2+5+8=33$. Thus, the answer is $45+33=\boxed{78}$.