Two nuclei have their mass numbers in the ratio of $1:3$. The ratio of their nuclear densities would be:

Nuclear density $\rho$ is given by $\rho =\frac{\text{Mass}}{\text{Volume}}$.Mass $M\approx A\times m_n$ (where $A$ is mass number).Volume $V=\frac{4}{3}\pi R^3=\frac{4}{3}\pi (R_0{A}^{1/3}{)}^3=\frac{4}{3}\pi R_0^{3}A$.
$$\rho =\frac{A\cdot m_n}{\frac{4}{3}\pi R_0^{3}A}=\frac{m_n}{\frac{4}{3}\pi R_0^3}$$
Notice that $A$ cancels out. This means nuclear density is approximately constant for all nuclei and independent of the mass number. Thus the ratio is $1:1$.