Calculate the nuclear radius of ${}^{125}\mathrm{Fe}$(in fermi) knowing that the nuclear radius of ${}^{27}\mathrm{Al}$ is 3.6 fermi

Nuclear radius  $R=R_0{A}^{1/3}\Rightarrow R\propto A^{1/3}$

For   $\mathrm{Al},\mathrm{A}=27,{\mathrm{R}}_{\mathrm{Al}}=3.6 \text{fermi, for }\mathrm{Fe},\mathrm{A}=125$

$\frac{{\mathrm{R}}_{\mathrm{Fe}}}{{\mathrm{R}}_{\mathrm{Al}}}=\frac{{\mathrm{A}}_{\mathrm{Fe}}}{{\mathrm{A}}_{\mathrm{Al}}}={\left(\frac{125}{27}\right)}^{1/3}\Rightarrow {\mathrm{R}}_{\mathrm{Fe}}=\frac{5}{3}{\mathrm{R}}_{\mathrm{Al}}=\frac{5}{3}\times 3.6 \text{fermi}=6.00 \text{fermi}$