Answer is (B)
Since, $n$ lists of each size $m$.
Since, each list is sorted in ascending order use directly merge procedure of merge sort algo.
Take two list and merge..so one pair will take $\mathbf{2m}$ time.
So, total pairs in first level will be $n/2$. So total cost for one level is $\mathbf{(n/2)*2m=nm}$.
In next level cost for one pair is $4m$ and no of pairs will be $n/4$.. so next level cost will be $nm$.
So, like this each level will have cost $nm$.
No of levels will be when we have one complete list..
$\mathbf{n/2^k =1}$..
$\mathbf{k=\log_2^ \ n}$.
So, total cost will be $\mathbf{ \log n *(nm)}$