Solution:

The first digit can be any number from $1$ to $5$, making $5$ possibilities. Each of the succeeding digits, from the ten-thousands digit to the tens digit, can be any of the six digits from $0$ to $5$.

Finally, we claim that there are exactly two possibilities for the last digit. Given the first five digits, if we append the digits $0$, $1$, and $2$ in turn, we get three consecutive integers, exactly one of which is a multiple of $3$. The same happens when we add the digits $3$, $4$, and $5$.

Thus the total number of multiples of $3$ is
$$5 \cdot 6 \cdot 6 \cdot 6 \cdot 6 \cdot 2 = 12960.$$