Answer
$$2.3 \times 10^{17} \ \mathrm{kg} / \mathrm{m}^{3}$$
Work Step by Step
For $^{209} \mathrm{Bi}$
$$\rho_{m}=\frac{M}{V}=\frac{0.209 \mathrm{kg} / \mathrm{mol}}{|4 \pi / 3| \ \left|1.2 \times 10^{-15} \mathrm{m}\left| \ \ |209|^{1 / 3} \ \ {^3}{| 5.02 \times 10^{23} / \mathrm{mol} |}\right. \right.} $$ $$=2.3 \times 10^{17} \mathrm{kg} / \mathrm{m}^{3}$$
