Two nuclei have mass numbers 64 and 125 . Assuming them to be...

The radius $R$ of a nucleus is related to its mass number $A$ by:
$R = R_{0}A^{\dfrac{1}{3}}$

The ratio of the radii of the two nuclei is:
$\dfrac{R_{1}}{R_{2}} = \left(\dfrac{A_{1}}{A_{2}}\right)^{\dfrac{1}{3}} = \left(\dfrac{64}{125}\right)^{\dfrac{1}{3}} = \dfrac{4}{5}$

Assuming the nuclei to be perfect spheres, the surface area $S$ is given by $S = 4\pi R^{2}$. Thus, the surface area is proportional to the square of the radius ($S \propto R^{2}$).

The ratio of their surface areas is:
$\dfrac{S_{1}}{S_{2}} = \left(\dfrac{R_{1}}{R_{2}}\right)^{2} = \left(\dfrac{4}{5}\right)^{2} = \dfrac{16}{25}$

Answer: $16:25$