Which arrangement represents the correct order of electron gain enthalpy (multiplied with negative sign) of the given atomic species?

The correct order of electron gain enthalpy (multiplied with negative sign) of the given atomic species is $\mathrm{Cl}<\mathrm{F}<\mathrm{S}<\mathrm{O}$.

In fluorine and chlorine. Due to the small size of F and the extremely strong electrical repulsion between the fluorine electrons, Cl has a high electron gain enthalpy. This makes the arrival of an additional electron less than ideal.
Due to its greater size and lower electron density in comparison to oxygen, sulfur has a higher electron gain enthalpy than oxygen.
So, $\mathrm{O}<\mathrm{S}<\mathrm{F}<\mathrm{Cl}$.
Cl has the largest negative electron gain enthalpy, whereas O has the least negative electron gain enthalpy. When taking into account the negative sign, the more negative value will be little and the less negative value will be high.
So, Cl<F<S<O will be the appropriate order.

Hence, option C is correct.