no of pages in logical address space=2^24/2^k=2^(24-k)
no of entries in page table= no of pages
no of frames =2^16/2^k=2^(16-k)
so 16 -k bits required to address frame
size of page table=2^24-k(16-k)/8
=2^(21-k)(16-k)
=2^(25-k)-k 2^(21-k)
answer is option a
