 Solution

A conventional body-centred cubic cell contains two atoms, so its edge $a$ obeys $\rho a^3=2(23m_u)$. Taking $m_u=1.66054\times 10^{-27}\mathrm{kg}$ gives $a\simeq 4.29\times 10^{-10}\mathrm{m}$. The neighbour shells of a body-centred cubic lattice have eight nearest neighbours that are at $\sqrt{3}a/2$; the six next-nearest neighbours lie along the cell axes. Thus the next-nearest-neighbour distance is $a\simeq 0.429\mathrm{nm}$.