Answer is 2MB.
We need a page table entry for each possible page (virtual) means for $2^{32}/4 KB = 2^{20}$ possible pages.
Each page table entry must address a physical page whose address can be of $\log (64 MB / 4KB) = 14 $bits.
We are mentioned 2 extra bits per page table entry- so 16 bits = 2 bytes per PTE.
So, size of page table = $2 \times 2^{20} = 2 \; MB.$