Answer is - c) 68.5 ns
Main memory and secondary memory are hierarchical. But as per convention page fault service time includes primary memory access time. Here we are not given page fault service time but secondary memory access time. So,
Effective Memory Access Time = $0.35 \times 10 + 0.65 \times (100+10) = 3.5 + 65+6.5 = 75.$
This is not among the choices. So, lets try excluding primary memory access time for secondary memory access:
Effective Memory Access Time = $0.35 \times 10 + 0.65 \times 100 = 3.5 + 65 = 68.5.$