b
Radius of the nucleus  \(R\,=\,R_{0} A^{1 / 3}\)

\(\therefore\)  \(\frac{R_{\mathrm{Al}}}{R_{\mathrm{Te}}}=\left(\frac{A_{\mathrm{Al}}}{A_{\mathrm{Te}}}\right)^{1 / 3}\)

Here , \(A_{\mathrm{Al}} =27, A_{\mathrm{Te}}=125, R_{\mathrm{Te}}=? \)

\(\frac{R_{\mathrm{Al}}}{R_{\mathrm{Te}}}=\left(\frac{27}{125}\right)^{1 / 3}=\frac{3}{5} \Rightarrow \quad R_{\mathrm{Te}}=\frac{5}{3} R_{\mathrm{Al}}\)