Answer is (A).
When we access $100$ distinct page in some order (for example $1,2,3\ldots 100$) then total number of page faults $=100$. At last, the $4$ page frames will contain the pages $100,99,98$ and $97$. When we reverse the string $(100,99,98,\ldots,1)$ then first four page accesses will not cause the page fault because they are already present in page frames. But the remaining $96$ page accesses will cause $96$ page faults. So, total number of page faults $=100+96=196$.