If the nucleus $\begin{array}{r}27\\ 13\mathrm{Al}\end{array}$ has nuclear radius of about 3.6 fm, then $\begin{array}{c}125\\ 32\end{array}T\mathrm{e}$ would have its radius approximately as

It has been known that a nucleus of mass number A has radius 

$R=R_0{A}^{1/3}$

and $A =$ mass number 

In case of ${ }_{13}^{27}\mathrm{Al}$ let nuclear radius be $R_1$

 and for $\begin{array}{l}125\mathrm{Te}\\ 32\end{array}$nuclear radius be $R_2$

For ${ }_{13}^{27}\mathrm{Al},{\mathrm{R}}_1={\mathrm{R}}_0(27{)}^{\frac{1}{3}}=3{\mathrm{R}}_0$

For ${ }_{32}^{125}\mathrm{Te},{\mathrm{R}}_2={\mathrm{R}}_0(125{)}^{\frac{1}{3}}=5{\mathrm{R}}_0$

$\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}=\frac{5{\mathrm{R}}_0}{3{\mathrm{R}}_0}=\frac{5}{3}{\mathrm{R}}_1=\frac{5}{3}\times 3.6=6\mathrm{fm}$