To solve this problem, we need to use the formula for the radius of a nucleus, which is given by:$\\ R = R_0 A^{1/3} \\$where $R_0$ is a constant and $A$ is the mass number of the nucleus.$\\$Given:$\\$• For $^{27}_{13}Al,$ the radius is $R_{Al} = R_0 \cdot 27^{1/3}. \\$• We need to find the radius of $^{125}_{53}Te. \\$The radius of $^{125}_{53}Te$ is given by:$\\ R_{Te} = R_0 \cdot 125^{1/3} \\$To find the relation between $R_{Te}$ and $R_{Al},$ we take the ratio:$\\ \frac{R_{Te}}{R_{Al}} = \frac{R_0 \cdot 125^{1/3}}{R_0 \cdot 27^{1/3}} \\$Simplify the expression:$\\ \frac{R_{Te}}{R_{Al}} = \left(\frac{125}{27}\right)^{1/3} \\$Calculate $\left(\frac{125}{27}\right)^{1/3}: \\ 125 = 5^3$ and $27 = 3^3 \\ \left(\frac{125}{27}\right)^{1/3} = \left(\frac{5^3}{3^3}\right)^{1/3} = \frac{5}{3} \\$Thus, $R_{Te} = \frac{5}{3} R_{Al}. \\$Therefore, the radius of $^{125}_{53}Te$ is nearly $\frac{5}{3} R_{Al}. \\$This corresponds to Option 1.