Let size of a page table entry be x bytes
Page Table Size = 4 KByte = 212B - requires 12 bits for addressing.
Logical Address Space = 264 B
Now, we know
Size of page table=(Logical Address Space/Page Size)*Page Table Entry Size
Size of 1st level page table = (264/212)*x
Size of 2nd level page table = (264/(212)2)*x2
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like that
Now, last level of page table fits into single page
So, (264/(212)y)*xy = 212
Using trial and error method and y=6
we get x6=224
then x=24B=16 B
Size of PTE=16B
and 6 level of page table are required