This might help.
The most confusing part is how there are loglogn levels, so here is my approach.
The question says we have logn sorted list, for me this was difficult to visualize.
so lets take it as “there are X sorted lists”, now we know if they were to be sorted
using merge sort a binary tree will be formed.
Now, that tree with X elements will have log(X) levels.
The question also gives us total number of elements that is $log(n)*n/log(n)= n$
So n comparisons at every level $n*log(X)$ replacing x by logn (as given in question)
O($nloglogn$)