The correct option is D C<N<F<O
The electronic configurations after first ionisation are given below:
C+→1s2 2s2 2p1
N+→1s2 2s2 2p2
O+→1s2 2s2 2p3
F+→1s2 2s2 2p4
The elements belong to the same period and the next electron to be withdrawn belongs to the 2p orbital in each case.
From C+ to F+ effective nuclear charge is increasing so I.E. also increases. But for F+ less energy is required than O+ because after losing one electron, the former will get a stable half-filled 2p3 configuration. On the other hand, more energy is required for removing electron from the already-stable half-filled 2p3 orbital for O+.