Answer
$$\rho_m = \boxed{2.3 \times 10^{17} \mathrm{kg} / \mathrm{m}^{3}}$$
Work Step by Step
As $V \propto r^{3}=\left(r_{0} A^{1 / 3}\right)^{3} \propto A,$
we find
$$\rho_{m} \propto A / V \propto A / A \approx \,\text{const}$$
This result is for all nuclides.
Hence, the mass density as in part (b)
$$\rho_m = \boxed{2.3 \times 10^{17} \mathrm{kg} / \mathrm{m}^{3}}$$
