To solve this problem, we need to compare the solubilities of two salts, MY and $NY_3,$ given their $K_{sp}$ values.$\\$The solubility product $K_{sp}$ is defined as the product of the concentrations of the ions each raised to the power of their coefficients in the balanced equation.$\\$For the salt MY, which dissociates as:$\\$MY$\rightleftharpoons$M$^+ +$Y$^- \\$The $K_{sp}$ expression is:$\\ K_{sp} = [$M$^+][$Y$^-] \\$Let $s$ be the molar solubility of MY. Then:$\\ [$M$^+] = s, \quad [$Y$^-] = s \\ K_{sp} = s \cdot s = s^2 \\ s^2 = 6.2 \times 10^{-13} \\ s = \sqrt{6.2 \times 10^{-13}} \\ s \approx 7.87 \times 10^{-7}$ M$\\$For the salt $NY_3,$ which dissociates as:$\\$NY$_3 \rightleftharpoons$N$^{3+} + 3$Y$^- \\$The $K_{sp}$ expression is:$\\ K_{sp} = [$N$^{3+}][$Y$^-]^3 \\$Let $s$ be the molar solubility of $NY_3.$ Then:$\\ [$N$^{3+}] = s, \quad [$Y$^-] = 3s \\ K_{sp} = s \cdot (3s)^3 = 27s^4 \\ 27s^4 = 6.2 \times 10^{-13} \\ s^4 = \frac{6.2 \times 10^{-13}}{27} \\ s = \left(\frac{6.2 \times 10^{-13}}{27}\right)^{1/4} \\ s \approx 1.24 \times 10^{-4}$ M$\\$Comparing the molar solubilities:$\\ s_{MY} \approx 7.87 \times 10^{-7}$ M$\newline s_{NY_3} \approx 1.24 \times 10^{-4}$ M$\\$Therefore, the molar solubility of MY in water is less than that of $NY_3. \\$This corresponds to Option 2.