option A is correct
no of stall stalls per memory access (memory reference)= Miss rate of L1 * Miss penalty of L1 +
Miss rate of L1 * Miss rate of L2 * Miss penalty of L2
=$(150/450)*30+((50/450)*(25/50))*60 =300/45$ stall
here 3 memory reference / instruction
so stall in one instructions=300*3/45=20