The transfer time should be $4*200 + 20 = 820$ ns. But this is not in option. So, I assume the following is what is meant by the question.
$L2$ block size being $16$ words and data width between memory and $L2$ being $4$ words, we require $4$ memory accesses(for read) and $4$ $L2$ accesses (for store). Now, we need to send the requested block to $L1$ which would require one more $L2$ access (for read) and one $L1$ access (for store). So, total time
$= 4 * (200 + 20) + (20 + 2)$
$= 880 + 22$
$= 902 \ ns$
Correct Answer: $C$