Process size = 512KB = $2^{19} B$
PTE = 4 Bytes
Page Size = 1024 Bytes = $2^{10} B$
Number of Pages needed = $2^{19}/2^{10} = 2^{9} = 512$
=$2^{6}*2^{10}=2^{16} = 64KB$ = 64KB*8 = 512KB
It means we need 8 3rd level page tables, each 3rd level contains 64 PT entries, and each entry points to 1 KB page
So we need 1 page table at first level, 1 page table at second level, 8 page tables at third level.
total memory required to store a 512KB size process:
$4*2^{8} + 4*2^{8} + 8*4*2^{6} = 2^{6}(16 + 16 + 32) = 64*64 = 2^{12}B = 4KB$
