Answer
$E_F = 9.43 eV$
Work Step by Step
The equation to find Fermi Energy is
$E_F = \frac{(0.121)(h)}{m_e} n^{2/3}$
Where
$h = 6.626\times 10^{-34} J.s$
$m_e=9.109\times 10^{-31} kg$
n from previous answer $n = 1.31 \times 10^{29} m^{-3}$
So straight away divide by charge $1.602 \times 10^{-19} J/eV$ to get answer in eV
$E_F = \frac{(0.121)((6.626\times 10^{-34} J.s)^2)}{(9.109\times 10^{-31} kg)(1.602 \times 10^{-19} J/eV)} (1.31 \times 10^{29} m^{-3})^{2/3}$
$E_F = 9.43 eV$
