A concentration cell at 298 K is set up as Ag | AgX(s), NaX (0.10 M) \| Ag^+ (0.10 M) | Ag, where AgX is a sparingly soluble salt. Its emf is 0.413…

The left electrode has $[\mathrm{Ag^+}]$ fixed by the solubility equilibrium, the right one has $0.10\ \mathrm{M}$. For a one-electron concentration cell $E=0.059\log\dfrac{0.10}{[\mathrm{Ag^+}]_L}$. So $\log\dfrac{0.10}{[\mathrm{Ag^+}]_L}=\dfrac{0.413}{0.059}=7.0$, giving $[\mathrm{Ag^+}]_L=1.0\times10^{-8}\ \mathrm{M}$. In the left compartment $[\mathrm{X^-}]=0.10\ \mathrm{M}$ (from NaX; the contribution of AgX is negligible), hence $K_{sp}=[\mathrm{Ag^+}][\mathrm{X^-}]=1.0\times10^{-8}\times0.10=1.0\times10^{-9}$. In pure water $s=\sqrt{K_{sp}}=\sqrt{10^{-9}}=3.2\times10^{-5}\ \mathrm{M}$. The option with $K_{sp}=10^{-8}$ forgets the $\mathrm{X^-}$ concentration (and then takes $s=\sqrt{K_{sp}}$); the other two pairs do not fit the measured emf. Note that the concentration cell has $E^\circ=0$, so only the ratio of the two silver ion concentrations matters and no standard potential data are needed.