Answer
$$1.0 \times 10^{25} \ \mathrm{C} / \mathrm{m}^{3}
$$
Work Step by Step
For $^{55} \mathrm{Mn}$, the charge density is
$$
\rho_{q}=\frac{Z e}{V}=\frac{(25)\left(1.6 \times 10^{-19} \mathrm{C}\right)}{(4 \pi / 3)\left[\left(1.2 \times 10^{-15} \mathrm{m}\right)(55)^{1 / 3}\right]^{3}}=1.0 \times 10^{25} \mathrm{C} / \mathrm{m}^{3}
$$
