Option $C$ is the answer. It can be done by Master's theorem.
$n^{\log_b a} = n^{\log_2 2} = n$.
$f(n) = \sqrt n = n^{\frac{1}{2}}$.
So, $f(n) = O\left(n^{\log_b a -\epsilon}\right)$ is true for any real $\epsilon$, $0 < \epsilon < \frac{1}{2}$. Hence Master theorem Case 1 satisfied,
$$T(n) = \Theta\left(n^{\log_b a}\right) = \Theta (n).$$