$L$ can be recognized by a dfa. we have a dfa to accept all such strings which when interpretated as decimal number are divisible by $n$. Where $n$ can be anything the dfa of such can be made by a trick.
States are equal to possible remainders
$$\overset{\text{Transition Table}}{\begin{array}{|c|c|c|c|}\hline
\textbf {} & \textbf {0} & \textbf{1} \\\hline
q_0 & q_0 & q_1 \\ \hline
q_1 & q_2 & q_3 \\\hline
q_2 & q_4 & q_0 \\\hline q_3 & q_1 & q_2 \\\hline q_4 &q_3 & q_4 \\\hline
\end{array}}$$
If you can see the symmetry in it. write states and make fill like $q_0 \ q_1 \ q_2 \ q_3\ q_4\ q_0 ...$
Now, it is saying that it has to always start with $1$ which the above dfa will not satisfy so make it a nfa by making a transition from $q_0$ on zero to a new dead state. now you have a nfa reduce it which will result in a deterministic DFA .
So, option is A.