= Solution
Interpret the sphere as an atom of <iron>, and assume neighbouring atoms touch without overlap. Its radius is $r=0.124\,\mathrm{nm}=1.24\times10^{-10}\,\mathrm m$, and its <mass> is $m_{\rm Fe}=56u\simeq9.30\times10^{-26}\,\mathrm{kg}$, using atomic mass unit $u=1.66054\times10^{-27}\,\mathrm{kg}$.
In the <face-centered cubic lattice>, atoms touch along a face diagonal: $\sqrt2a_F=4r$, so $a_F=2\sqrt2r=0.3507\,\mathrm{nm}$. Four atoms occupy a conventional cube. The <hard-sphere atomic volume in cubic lattices> therefore gives
$$
\boxed{\Omega_F=\frac{a_F^3}{4}=4\sqrt2r^3=1.08\times10^{-29}\,\mathrm{m^3},\qquad
\rho_F=\frac{56u}{\Omega_F}=8.62\times10^3\,\mathrm{kg\,m^{-3}}.}
$$
In the <body-centered cubic lattice>, atoms touch along the body diagonal: $\sqrt3a_B=4r$, so $a_B=4r/\sqrt3=0.2864\,\mathrm{nm}$. There are two atoms per conventional cube, giving
$$
\boxed{\Omega_B=\frac{a_B^3}{2}=\frac{32r^3}{3\sqrt3}=1.17\times10^{-29}\,\mathrm{m^3},\qquad
\rho_B=7.92\times10^3\,\mathrm{kg\,m^{-3}}.}
$$
These are <primitive unit cell> volumes, not the volumes of the conventional cubes or of the atomic spheres. The denser FCC contact packing gives the higher <mass density> in this fixed-radius model. Actual hot phases have temperature-dependent dimensions and bonding, so these numbers should not be treated as exact experimental phase densities.
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