The outer loop is running for $n/2$ times and inner loop is running for $\log_2 n$ times (each iteration doubles $j$ and $j$ stops at $n$ means $\log_2 n$ times $j$ loop will iterate).
Now in each iteration $k$ is incremented by $n/2$. So, overall $k$ will be added $n/2 \ast \log n \ast n/2$ with an initial value of $0$. So, final value of $k$ will be $\Theta(n^2 \log n)$.
Correct Answer: B.