Page table is stored in main memory. Hence we need a frame number to access the page table. Frame number tells the address of a page in main memory.
Now first focus on the VA. It has 32 bits out of which 12 least significant bits are for offset. So Size of page = 4KB.
PTE size = 4 Bytes. So one page table contains $2^{10}$ entries.
Now look at the VA division. It is divided as 2 | 9 | 9 | 12
So third level page table has the address where the actual frame is loaded. Hence it will require 24 bits.
2nd level PT points to the next level page table. And also one page has $2^{10}$ entries and we are addressing $2^{9}$ entries so one page has 2 page tables. So 1 extra bit is needed to distinguish between the page tables. Hence 25 bits needed. Same for the next level.
Here, One page has two page tables.