Since $L$ is in NP it is decidable (recursive) and so is its complement $L^c$. Now, $L^c$ may or may not be in NP. But we are given that for any string length $n$, exactly one string belong to $L$, which means for any string length all but one string belong to $L^c$.
Now, definition of NP says that all instances of the problem can be solved in polynomial time using a nondeterministic TM. So, given an instance of $\langle L^c, x\rangle$, we non-deterministically take all words of length $n$, where $n$ is the length of $w$, and see if it is in $L$. As soon as we get the word (such a word is sure to exist as exactly one word of length $n$ belong to $L$), we see if this word is same as $x$. If it is not same (and only if it is not same), $x \in L^c$ and we get this answer in polynomial time making $L^c$ an NP problem.