A 2-level paging scheme is used.
It is already given in the question that the applicable page table is divided into 2K pages
=> 1st level page table do not fit in a single frame. Because if it fits into one frame then we will not get 2k pages of size 4KB.
=> in 2nd level page table, #entries = #pages created for 1st level page table = 2K, and so the answer.
Physical address calculation:
Given page table is divided into pages of size 4K and we know in case of paging, page-size and frame-size is same throughout.
So offset is 12-bit.
Also 16k frames are there => 14bit required to access each frame uniquely.
PA: 14|12
Logical Address Calculation:
Offset is 12 bit(calculated above).
2nd level page has 2k entries => 12 bit
1st level page we need to calculate. Given 1st level has 2K pages each of size 4KB => 1st level Page table size is 2K*4KB = 2000 * 4KB
Also page table entry is 2 bytes => #entries in 1st level page table = 2000 * 4KB / 2B = 4M(approx) => 22 bit required
So logical address: 12|22|12