Second ionisation potential of oxygen is

The Electronic configuration of oxygen can be represented as $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^4$.

After Ist ionisation it becomes $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^3$and after IInd ionisation It becomes $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^2$. 

It means it is less than that of fluorine.

Hence,option(B) is the answer.