Answer
$$8.8 \times 10^{24} \ \mathrm{C} / {\mathrm{m}}^{3} $$
Work Step by Step
For $^{209}$ Bi, the charge density is
$$\rho_{q}=\frac{Z e}{V}=\frac{|83|\ \ |1.6 \times 10^{-19} \mathrm{C} |}{|4 \pi / 3|\ \ |1.2 \times10^{-15} \mathrm{m}| \ \ {| 209|^{1 / 3}}^{3}}=8.8 \times 10^{24} \mathrm{C} / {\mathrm{m}}^{3} $$
