(c)

Given, nuclear radius is

$R=r_0 A^{\frac{1}{3}}$

Here, atomic mass number of nucleus $=A$

$\therefore$ Nuclear density $d$ is given by

$d=\frac{\text { Mass number }}{\text { Volume }}$

$\Rightarrow \quad d=\frac{A}{\frac{1}{3} \pi R^3}=\frac{4}{3} \pi\left(r_0 A^{\left.\frac{1}{3}\right)^3}\right.$

$\Rightarrow \quad d=\frac{A}{\frac{1}{3} \pi r_0^3 \cdot A}=\frac{3}{4 \pi r_0^3}$

As $r_0=$ a constant, so nuclear density is a constant quantity.

$\therefore$ Nuclear mass density of $U^{238}$ is same as that of $Sn ^{119}$.