Considering memory is byte addressable.
Given VA = 64 bit
So, Virtual Address Space (VAS) = $2^{64} bytes$
Given page offset is 16 bits, so page size = $2^{16} bytes$
Lets take page table entry size as $x$ Bytes
Now first level paging is done, $\frac{2^{64} }{2^{16}} = 2^{48}$
So first level page is divided into $2^{48}$ pages and size of each PTE is $x$ bytes so total size of first level page is $2^{48}\times x$ bytes
Now we are applying paging for second time
$\frac{2^{48}\times x}{2^{16}} = 2^{32}$
$x = 1$
So page table entry size if 1B
So second level page size = $2^{32}\times 1$ bytes i.e 4GB which is option a
But there is error in the question, even we can not fit second level page table into one page, since its size is bigger than page size. So we need to go for another level, but in question they have mention only two level paging has been done which is wrong.