To solve this problem, we need to use the formula for the radius of a nucleus,$\\ R = R_0 A^{1/3} \\$where $R_0$ is a constant and $A$ is the mass number of the nucleus.$\\$Given:$\\$• The mass number of the first daughter nucleus $A_1 = 125. \\$• The mass number of the second daughter nucleus $A_2 = 64. \\$The radii of the daughter nuclei are given by:$\\ R_1 = R_0 A_1^{1/3} \\ R_2 = R_0 A_2^{1/3} \\$We need to find the ratio of the radii $\frac{R_1}{R_2}. \\$Substitute the expressions for $R_1$ and $R_2: \\ \frac{R_1}{R_2} = \frac{R_0 A_1^{1/3}}{R_0 A_2^{1/3}} \\$The $R_0$ terms cancel out:$\\ \frac{R_1}{R_2} = \frac{A_1^{1/3}}{A_2^{1/3}} \\$Substitute the given mass numbers:$\\ \frac{R_1}{R_2} = \frac{125^{1/3}}{64^{1/3}} \\$Calculate the cube roots:$\\ 125^{1/3} = 5 \\ 64^{1/3} = 4 \\$Thus, the ratio is:$\\ \frac{R_1}{R_2} = \frac{5}{4} \\$Therefore, the ratio of the radius of the two daughter nuclei is $5 : 4. \\$This corresponds to Option 3.