Lets say ,page size = 2^x byte.
so no of bits required for offset =x.
now page table entry = 4 B
now ,max no of entry one page can store is 2^x/4 =2^(x-2).
so each inner and outer level page table fit completely in one page frame . so we can write,
outer level inner level offset,
now it sums up to virtual address space .2^38.
(x-2)+(x-2)+x=38
x=14.
so page size =16 KB