If radius of the ${}_{13}{}^{27}\mathrm{Al}$ nucleus is taken to be $R_{\mathrm{Al}}$, then the radius of ${}_{53}{}^{125}\mathrm{Te}$ nucleus is nearly

$\text{ Radius of the nucleus }R=R_0{A}^{1/3}$

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

$\text{ Here, }{A}_{\mathrm{Al}}=27,A_{\mathrm{Te}}=125,R_{\mathrm{Te}}=?$

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