Given page table entry size is 4 bytes = 22 B
the memory required to store all the page tables is 2^a(2^b-1) bytes .
32 bit virtual addresses.
page size = 4KB = 22 * 210 B = > offset bits = 12
Number of entry in inner page table = page size / page table entry size
= 212 / 22
= 210
outer page table will require 32-12(page offset)-10(inner page table)= 10 bits
so for outer page tables there will be 210 entries, for inner page tables there will be 210*210 entries
For 20 level paging , total entires including all the page tables= 21+22+23+24.......220
=2*(220-1) entries,
now each entry is of 4 bytes so total size = 2*(220-1)*4
memory required to store all the page tables = (220-1) * 23
=> 2a * (2b-1) = (220 - 1 ) * 23
a= 3 and b = 20
hence a*b = 3 * 20 = 60