For purpose of solving I am assuming that n=2$^{c}$ where c>1
1) outer loop or i$^{th}$ loop will execute (log n) times
2) middle loop or j$^{th}$ loop will execute (log n)+1 times
3) now for last loop or K$^{th}$ loop will run for :
$\sum_{k=1}^{log n+1} \left ( \frac{n}{2}+1-k \right )$
Eg:
for n=8, 1$^{st}$ loop will execute 3 times
now 2$^{nd}$ will execute (log n +1) i.e 4 times
and 3$^{rd}$ loop will execute 10 times as per formula
we can simplify $\sum_{k=1}^{log n+1} \left ( \frac{n}{2}+1-k \right )$ as = $\frac{n}{2} - \left ( 1+2+3...log n \right )$
since -(1+2+3...log n ) is negligible we will omit in result
so final answer is (log n)*((log n)+1)*($\frac{n}{2}$)= O(n* (log n)$^{2}$)
Answer is B