Step 1 — Extract limiting molar conductivities using Debye-Huckel-Onsager form ${\Lambda }_m={\Lambda }_m^{\circ }-b\sqrt{C}$.

For each salt, the pair of $(C,{\Lambda }_m)$ values gives two equations:

NaNO${}_3$: $111={\Lambda }^{\circ }-0.1b,101={\Lambda }^{\circ }-0.2b\Rightarrow {\Lambda }_{{\text{NaNO}}_3}^{\circ }=121$ S cm${}^2$ mol${}^{-1}$.

NaCl: $117={\Lambda }^{\circ }-0.1b,107={\Lambda }^{\circ }-0.2b\Rightarrow {\Lambda }_{\text{NaCl}}^{\circ }=127$ S cm${}^2$ mol${}^{-1}$.

AgNO${}_3$: $125={\Lambda }^{\circ }-0.1b,116={\Lambda }^{\circ }-0.2b\Rightarrow {\Lambda }_{{\text{AgNO}}_3}^{\circ }=134$ S cm${}^2$ mol${}^{-1}$.

Step 2 — Kohlrausch's law for AgCl.

${\Lambda }_{\text{AgCl}}^{\circ }={\Lambda }_{{\text{AgNO}}_3}^{\circ }+{\Lambda }_{\text{NaCl}}^{\circ }-{\Lambda }_{{\text{NaNO}}_3}^{\circ }=134+127-121=140$ S cm${}^2$ mol${}^{-1}$.

Step 3 — Relate conductivity to solubility. Since the saturated solution is taken to be at the limiting concentration,

${\Lambda }_{\text{AgCl}}^{\circ }=\frac{\kappa \times 1000}{S}\Rightarrow 140=\frac{(1.40\times 10^{-6})(1000)}{S}$.

$S=X=10^{-5}$ mol L${}^{-1}$, so ${\log }_{10}(X^{-1})=5$, matching option (C).