PAGING

Question

Comments

how could no. of bits for outer page table greater than inner page table??

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Incomplete question

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Yes I feel that too,Is there any way from above given information we could get PTE size.

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C is out of options

Minimum possible page table entries size would be 1byte so pte would be 4GB

If pte is 2 byte then 8Gb and if pte is 16 byte then 32GB

Answer 1

Considering memory is byte addressable.
Given VA = 64 bit

32

16

16

So, Virtual Address Space (VAS) = $2^{64} bytes$
Given page offset  is 16 bits, so page size = $2^{16} bytes$
Lets take page table entry size as $x$ Bytes

Now first level paging is done, $\frac{2^{64} }{2^{16}} = 2^{48}$
So first level page is divided into $2^{48}$ pages and size of each PTE is $x$ bytes so total size of first level page is $2^{48}\times x$ bytes

Now we are applying paging for second time
$\frac{2^{48}\times x}{2^{16}} = 2^{32}$
$x = 1$
So page table entry size if 1B

So second level page size  = $2^{32}\times 1$ bytes i.e 4GB which is option a

But there is error in the question, even we can not fit second level page table into one page, since its size is bigger than page size. So we need to go for another level, but in question they have mention only two level paging has been done which is wrong.