The nuclear radius $R$ is related to the mass number $A$ by the relationship $R = R_0 A^{1/3}$, where $R_0$ is a constant (approximately 1.2 fm).

For Aluminum ($Al$), the mass number $A_1 = 27$ and radius $R_1 = 3.6$ fm. For Tellurium ($Te$), the mass number $A_2 = 125$.

Taking the ratio: $\frac{R_2}{R_1} = \left(\frac{A_2}{A_1}\right)^{1/3}$ $\frac{R_2}{3.6} = \left(\frac{125}{27}\right)^{1/3}$ $\frac{R_2}{3.6} = \frac{5}{3}$ $R_2 = 3.6 \times \frac{5}{3} = 1.2 \times 5 = 6.0$ fm.