Density of nuclear matter is independent of mass number, so the required ratio is 1: 1. Alternative
$A_{1}: A_{2}=1: 3$
Their radii will be in the ratio
$R_{0} A_{1}^{1 / 3}: R_{0} A_{2}^{1 / 3}$
$=1: 3^{1 / 3}$
Density $=\frac{A}{\frac{4}{3} \pi R^{3}}$
$\therefore \rho_{A_{1}}: \rho_{A_{2}}=\frac{1}{\frac{4}{3} \pi R_{0}^{3} \cdot 1^{3}}$
$=\frac{4}{3} \pi R_{0}^{3}\left(3^{1 / 3}\right)^{3}$
Their nuclear densities will be the same.