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a simple formula :- 

let there are 'n' levels in paging.

last level page table size = $\LARGE\frac{(\;process\;size\;) \;*\; (\; PTE\;)^{n}}{ (\; Page\;Size\;)^{n}}$

given that last level page table exactly fit in one Page.

 

==>  $\LARGE\frac{(\;process\;size\;) \;*\; (\; PTE\;)^{n}}{ (\; Page\;Size\;)^{n}}$ = Page Size

process size = 246 B

PTE = 32 bits = 4 Bytes = 22 B

Page Size = 2P B

n = 3

substitute you get P = 13 ===> Page Size = 213 B = 8 KB

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