A saturated solution of silver chloride in the laboratory water has a conductivity of 2.20 × 10⁻⁴…

Step 1. The salt's conductivity. $\kappa = 2.20 \times 10^{-4} - 0.40 \times 10^{-4} = 1.80 \times 10^{-4}\ \text{S/m}$.

Step 2. The limiting value. $\Lambda_m^\circ = 61.9 + 76.3 = 138.2\ \text{S cm}^2/\text{mol} = 1.382 \times 10^{-2}\ \text{S m}^2/\text{mol}$.

Step 3. The solubility. $c = \dfrac{\kappa}{\Lambda_m^\circ} = \dfrac{1.80 \times 10^{-4}\ \text{S/m}}{1.382 \times 10^{-2}\ \text{S m}^2/\text{mol}} = 1.30 \times 10^{-2}\ \text{mol/m}^3$. Divide by $1000$ to get $1.30 \times 10^{-5}\ \text{mol/L}$.

Step 4. The solubility product. Each formula unit gives one $\text{Ag}^+$ and one $\text{Cl}^-$, so $K_{sp} = c^2 = (1.30 \times 10^{-5})^2 = 1.7 \times 10^{-10}\ \text{mol}^2/\text{L}^2$.