= Solution
Count two <electron spin> states in each reciprocal-space cell of volume $(2\pi)^3/V$. Filling a <Fermi sphere> at zero <temperature> gives
$$
n=\frac{2}{(2\pi)^3}\frac{4\pi k_F^3}{3},\qquad
k_F=(3\pi^2n)^{1/3}=1.360\times10^{10}\,\mathrm{m^{-1}}.
$$
The <free electron model> then gives
$$
\boxed{E_F=\frac{\hbar^2}{2m_e}(3\pi^2n)^{2/3}
=1.129\times10^{-18}\,\mathrm J\simeq7.05\,\mathrm{eV}.}
$$
The <energy> zero is the bottom of the model's <conduction band>. The supplied <carrier density> already incorporates the one-electron-per-atom assumption; it must not be multiplied by another atomic count.
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