The order of first electron affinity of 0, S and Se is :

The order of first electron affinity of 0, S and Se is : $\mathrm{S}>\mathrm{Se}>\mathrm{O}$.

O's electron affinity $\left(-141\mathrm{kJ}{\mathrm{mol}}^{-1}\right)$ value is incredibly low due to its lower atomic size than sulfur (weaker electron-electron repulsion in 3p-subshell). It is also less than Se and Te.
$\mathrm{S}>\mathrm{Se}>\mathrm{O}$.

Hence, option B is correct.