In weakly alkaline medium $\mathrm{MnO}{}_4{}^{-}$ oxidises $\mathrm{I}{}^{-}$ to $\mathrm{IO}{}_3{}^{-}$ and is itself reduced to $\mathrm{MnO}{}_2$. The mass of $\mathrm{KMnO}{}_4$ (molar mass $158\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$) required to oxidise $0.05$ mol of KI in this medium is

Mn changes from +7 to +4 (3 electrons per $\mathrm{MnO}{}_4{}^{-}$) while I changes from $-1$ to +5 (6 electrons), so 2 mol $\mathrm{MnO}{}_4{}^{-}$ per mol $\mathrm{I}{}^{-}$: $2\mathrm{MnO}{}_4{}^{-}+\mathrm{I}{}^{-}+\mathrm{H}{}_2\mathrm{O}\to 2\mathrm{MnO}{}_2+\mathrm{IO}{}_3{}^{-}+2\mathrm{OH}{}^{-}$ (balanced in atoms and charge). Moles = $0.05\times 2=0.10$, mass = $0.10\times 158=15.8\mathrm{g}$. The 9.48 g option wrongly uses the acidic 5-electron change.