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