Worst-case number of arithmetic operations performed by recursive binary search on a sorted array is given by the following recurrence relation:
$T(n) = T(\frac{n}{2}) + \Theta(1) $
$\text{for }~T(n) = aT(\frac{n}{b}) + f(n), \\ a\geq 1, b > 1$
$\text{Case 2 of master theorem says,}$
$\text{if}~~f(n) = \Theta(n^{\log_ba}) \\ \text{then}~~T(n) = \Theta(n^{\log_ba}\cdot \log(n)) $
$\text{Here, }n^{\log_b(a)} = n^{\log_2(1)} = n^0 = 1\\ \Rightarrow f(n) = \Theta(n^{\log_b(a)}) $
$\text{Hence, we can conclude that }T(n) = \Theta(1\cdot \log_2(n)) = \Theta(\log_2(n))$
B is correct