If radius of the $Al _{13}^{27}$ nucleus is estimated to be 3.6 Fermi then the radius of $Te_{52}^{125}$ nucleus be nearly

4 fermi  5 fermi 6 fermi 8 fermi

6 fermi

$ R = R_0 A^{1/3}$ $\frac{R_2}{R_1} = {\frac{A_2}{A_1}}^{1/3}$ $ R_2 = 3.6 \times \frac{5}{3} = 6 fermi$