For a saturated solution of $AgCl$ at $25^{\circ} C , k =3.4 \times 10^{-6} ohm ^{-1} cm ^{-1}$ and that of $H _{2} O (\ell)$ used is $2.02 \times 10^{-6} ohm ^{-1} cm ^{-1} \cdot \lambda_{m}^{0}$ for $AgCl$ is $138 ohm ^{-1} cm ^{2} mol ^{-1}$, then the solubility of $AgCl$ in moles per liter will be:
(A) $10^{-5}$
(B) $10^{-10}$
(C) $10^{-14}$
(D) $10^{-16}$