Answer :
The page table entry is 64 bits due to the utilization of 64-bit addressing. With the outer page table containing 18 bits, it can point to 2^18 pages, resulting in 2^18 entries.
Given that the page table entry corresponds to 64 bits due to the utilization of 64-bit addressing, and the outer page table encompasses 18 bits, it implies that it can reference 2^18 pages, leading to 2^18 entries.
Consequently, the calculation for the total page table size involves multiplying the number of entries (2^18) by the size of each entry (64 bits), yielding 2^18 * 64 bits.
Converting bits to bytes (8 bits per byte), we get 2^18 * 8 bytes, which further simplifies to 2MB.
