Calculate the molar solubility of silver chloride (AgCl) given its Ksp value.

When AgCl dissolves in water, it dissociates into silver ions ($\text{Ag}^+$) and chloride ions ($\text{Cl}^-$).

$$\text{AgCl(s)} \rightleftharpoons \text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$$

The solubility product constant for AgCl is given by the product of the concentrations of its ions at equilibrium. If 's' represents the molar solubility of AgCl, then $[\text{Ag}^+] = \text{s}$ and $[\text{Cl}^-] = \text{s}$.

$$\text{Ksp} = [\text{Ag}^+][\text{Cl}^-] = (\text{s})(\text{s}) = \text{s}^2$$

Given $\text{Ksp} = 2.0 \times 10^{-10} \text{ mol}^2\text{dm}^{-6}$.

$$\text{s}^2 = 2.0 \times 10^{-10} \text{ mol}^2\text{dm}^{-6}$$ $$\text{s} = \sqrt{2.0 \times 10^{-10} \text{ mol}^2\text{dm}^{-6}}$$ $$\text{s} = 1.4142 \times 10^{-5} \text{ mol dm}^{-3}$$

The quantity of silver lost in the process refers to the concentration of $\text{Ag}^+$ ions that dissolve in the water. This concentration is equal to the molar solubility (s).
Quantity of silver lost = $[\text{Ag}^+] = \text{s} = 1.414 \times 10^{-5} \text{ mol dm}^{-3}$.

The value matches option D.

The final answer is $\text{D}$.