If the nuclear radius of ${ }^{27}\mathrm{Al}$ is 3.6 Fermi, the approximate nuclear radius of ${ }^{64}\mathrm{Cu}$ in Fermi is: [2012] 

The radius of the nucleus is directly proportional to cube root of atomic number i.e., $R\propto A^{1/3}$$\Rightarrow R=R_0{A}^{1/3}$, where $R_0$ is a constant of proportionality

$\frac{R_2}{R_1}={\left(\frac{A_2}{A_1}\right)}^{\frac{1}{3}}={\left(\frac{64}{27}\right)}^{\frac{1}{3}}=\frac{4}{3}$

where $R_1$ = the radius of ${ }^{27}\mathrm{Al}$, and $A_1$ = Atomic mass number of Al

$R_2$ = the radius of $64\mathrm{Cu}$ $A_2$ = Atomic mass number of Cu 

${\mathrm{R}}_2=3.6\times \frac{4}{3}=4.8 fermi$