virtual address = 52 bits
page size = 16kB = 14 bits
no. of pages = 238
hence sum of all bits (in ans) = 38
also for frames ,
no. of frames = 32-14=218
page table entry = 18+ (additional) 14 bits = 32 bits = 22B
in each page = 214/22=212
i.e. at max there can be 212 pages in any page table .
hence, ans is option A)
< 2, 12, 12, 12 >