The radius of germanium (Ge) nuclide is measured to be twice the radius of ${ }_4^{9}Be$ . The
number of nucleons in Ge will be

The quantity of nucleons in an atom's nucleus determines the radius of the nucleus. The nucleus is larger the higher the number. Particularly, the radius is directly proportional to the mass number's cube root (which is the same as the nucleon number). Hence
$R\propto A^{\frac{1}{3}}\Rightarrow R=k{A}^{\frac{1}{3}}$ 
From the equation above,
$\Rightarrow \frac{R}{A^{{\displaystyle \frac{1}{3}}}}=k$
As this constant is same for all atoms, 

$\Rightarrow \frac{R_{Ge}}{A_{Ge}^{{\displaystyle \frac{1}{3}}}}=\frac{R_{Be}}{A_{Be}^{{\displaystyle \frac{1}{3}}}}$
Rearranging, we have that
$\Rightarrow \frac{R_{Ge}}{R_{Be}}=\frac{A_{Ge}^{{\displaystyle \frac{1}{3}}}}{A_{Be}^{{\displaystyle \frac{1}{3}}}}$
 $\Rightarrow \frac{R_{Ge}}{R_{Be}}={\left(\frac{A_{Ge}}{A_{Be}}\right)}^{\frac{1}{3}}$
Cubing both sides, we have that
$\Rightarrow \frac{R_{Ge}^3}{R_{Be}^3}=\frac{A_{Ge}}{A_{Be}}$
Making $A_{Ge}$ subject of the formula, 

 $\Rightarrow A_{Ge}={\left(\frac{R_{Ge}}{R_{Be}}\right)}^3{A}_{Be}$
The radius of germanium, according to the question, is twice that of beryllium, therefore, $R_{Ge}=2R_{Be}$
Hence, we get
 $\Rightarrow A_{Ge}={\left(\frac{2R_{Be}}{R_{Be}}\right)}^3{A}_{Be}$
Cancelling $R_{Be}$ and substituting the value of mass number of  ${}_{4}B{e}^9$
$\Rightarrow A_{Ge}={\left(2\right)}^3\times 9$
Thus, $A_{Ge}=8\times 9=72nucleons$
Hence the correct answer is 72.