Answer: Option C
Number of pages = 232 / 4KB = 220 as we need to map every possible virtual address.
So, we need 220 entries in the page table. Physical memory being 64 MB, a physical address must be 26 bits and a page (of size 4KB) address needs 26-12 = 14 address bits. So, each page table entry must be at least 14 bits.
So, total size of page table = 220 * 14 bits ~ 2 MB