Answer
$E_F = 5.49 eV$
Work Step by Step
The Fermi Energy Equation
$E_F = \frac{(0.121)(h^2)}{m_e} n^{2/3}$
where
$h = 6.626\times 10^{-34} J.s$
$m_e=9.109\times 10^{-31} kg$
n from previous answer $n=5.86 \times 10^{28} m^{-3}$
so
$E_F = \frac{(0.121)((6.626\times 10^{-34} J.s)^2)}{9.109\times 10^{-31} kg} (5.86 \times 10^{28} m^{-3})^{2/3}$
$E_F =8.80 \times 10^{-19} J$
Convert to eV
$E_F = 5.49 eV$
