At 298 K the following concentration cell is set up: Ag(s) | AgCl(s), NaCl(aq, 0.010 M) \| AgNO3(aq, 0.10 M) | Ag(s). The measured emf is 0.413 V.…

Both electrodes are silver, so the cell is a concentration cell and only the $\mathrm{Ag^+}$ activities matter. The left half-cell has the smaller $[\mathrm{Ag^+}]$, hence it is the anode, and with one electron transferred $E=0.059\log\dfrac{[\mathrm{Ag^+}]_{R}}{[\mathrm{Ag^+}]_{L}}$. Putting $E=0.413$ gives $\log\dfrac{0.10}{[\mathrm{Ag^+}]_L}=7$, so $[\mathrm{Ag^+}]_L=1.0\times10^{-8}\ \mathrm{M}$. This silver ion concentration is fixed by the solubility equilibrium in the presence of $0.010\ \mathrm{M\ Cl^-}$, so $K_{sp}=[\mathrm{Ag^+}][\mathrm{Cl^-}]=10^{-8}\times10^{-2}=1.0\times10^{-10}$. Option (A) is the silver ion concentration alone, a common slip when the chloride is forgotten. Option (C) results from treating the solution as saturated in pure water, $K_{sp}=s^2=(10^{-8})^2$, and (D) is not obtainable from the data. The answer is $1.0\times10^{-10}$.