Suppose page size is 2x B
Level 1: 2x B =( 32B / 2x B ) * 4B
2x = 232 - X + 2
Level 2 : 2x B = [ ( 232 - X + 2 ) / 2x B ] * 4B
2x = 232 - 2X + 2 * 2
So x =12 . PageSize = 212B
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Let the page size = 2x B
$\therefore$ Total pages in virtual memory
= VA space / page size
= $\frac{4 * 2^{30}}{2^{x}}$
= $2^{32-x}$
$\therefore$ Total entries in the innermost page table level
= $2^{32-x}$
Now, if 1 level of paging is used:
Total entries in the innermost page table level * PTE size = page size
$\therefore$ $2^{32-x}$ * 4B = $2^{x}$
$\therefore$ x=17
Now, if 2 level of paging is used:
Total pages in the innermost level of paging
= Total entries in the innermost page table level * PTE size / page size
= $2^{34-2x}$
Thus, size of outermost page table
= Total pages in the innermost level of paging * PTE size
= $2^{36-2x}$
$\therefore$ size of outermost page table = page size
$\therefore$ x=12
This, minimum page size = 212B