The major product of the following reaction is

The substrate contains two possible leaving groups: an alkyl iodide and an aryl fluoride.

Sodium thiophenoxide $$\mathrm{(PhS^-Na^+)}$$ acts as a strong nucleophile, and DMF is a polar aprotic solvent that strongly favors $$\mathrm{S_N2}$$ reactions.

The alkyl iodide undergoes substitution much more readily because $$\mathrm{I^-}$$ is an excellent leaving group, whereas nucleophilic aromatic substitution on the fluoronitrobenzene ring is comparatively less favorable under these conditions.

Thus, $$\mathrm{PhS^-}$$ attacks the carbon bearing iodine via backside attack through an $$\mathrm{S_N2}$$ mechanism.

Since $$\mathrm{S_N2}$$ reactions proceed with inversion of configuration, the incoming $$\mathrm{-SPh}$$ group appears on the side opposite to the original iodine substituent.

The aromatic ring and its substituents remain unchanged.

Hence, the correct product is the one showing $$\mathrm{-SPh}$$ on a dashed bond due to inversion of stereochemistry.