Answer will be $O(log(n^{2}))$.
Here, $log(n^{2}) = 2logn$ which leads to $O(2logn) = O(logn)$ as you drop the constants in Big-Oh notation.
So, both A and C.
Note, too, that $O(log(n))$ is exactly the same as $O(log(n^{c}))$. The logarithms differ only by a constant factor, and the big O notation ignores that. Similarly, logs with different constant bases are equivalent.