Explanation:
B It has been found that a nucleus of mass number $A$ has a radius $R=R_{0} A^{1 / 3}$
Where, $R_{0}=1.2 \times 10^{-15} \mathrm{~m}\left(=1.2 \mathrm{fm} ; 1 \mathrm{fm}=10^{-15} \mathrm{~m}\right)$. This means the volume of the nucleus, which is proportional to $\mathrm{R}^{3}$ is proportional to $\mathrm{A}$. Thus the density of nucleus is a constant, independent of $\mathrm{A}$, for all nuclei. The density of nuclear matter is approximately $2.3 \times 10^{17} \mathrm{~kg} \mathrm{~m}^{-3}$.
So, ratio of density of oxygen nucleus to helium nucleus,
$\rho_{\mathrm{O}}=\rho_{\mathrm{He}}$
$\frac{\rho_{\mathrm{O}}}{\rho_{\mathrm{He}}}=\frac{1}{1}$
$\rho_{\mathrm{O}}: \rho_{\mathrm{He}}=1: 1$