EDIT:
As upon insertion of the k-th element, 2k-th element, 3k-th element and so on, we take additional (k) steps.
So for insertion of K^33 elements no of sets of (k) elements:
k+k+k+.......................... +k (n terms)= k^33
k*n=k^33 => n=k^32
As given each set of k insertion takes (k) steps .
So total cost of insertion for k^33 elements= no of set of k insertion * cost per set insertion = k^32 * k = ()
so cost per operation = (1)
so ans should be option (a)