Answer
$$2.3 \times 10^{17} \mathrm{kg} / \mathrm{m}^{3}
$$
Work Step by Step
For $^{55} \mathrm{Mn}$ the mass density is
$$
\rho_{m}=\frac{M}{V} $$ $$=\frac{0.055 \mathrm{kg} / \mathrm{mol}}{(4 \pi / 3)\left[\left(1.2 \times 10^{-15} \mathrm{m}\right)(55)^{1 / 3}\right]^{3}\left(6.02 \times 10^{23} / \mathrm{mol}\right)}=2.3 \times 10^{17} \mathrm{kg} / \mathrm{m}^{3}
$$
