If number is divisible by 6 , then it should be an even number and divisible by 3 also, so sum of the digits should be divisible by 3 .

Now last digit can be fill by 0 or 2 or 4 . First place can be filled by 5 choices (except 0 ) second and third can be filled by 6 choices.

Now fourth place is filled such that it must be of form $3 k$ or $3 k+1$ or $3 k+2$. So it can be fill by 2 choices ( 0 or 3 ) as whole number must be divisible by 3 so, total numbers
$$
\begin{aligned}
& =5 \times 6 \times 6 \times 2 \times 3 \\
& =1080
\end{aligned}
$$

Hence, option (d) is correct.