Answer
$E_F = 0.744 eV$
Work Step by Step
If measured from the top of the valence band, the energy of donor state is
$E = 1.11 \space eV - 0.11 \space eV = 1.0 \space eV$
Solve for $E_F$ taking the Fermi Energy formula
$ E = E_F + kT ln (P^{-1} - 1)$
$ E_F = E - kT ln (P^{-1} - 1)$
Where
$k = 8.62 \times 10^{-5} eV/K$
$T = 300K$
Probability occupied by electron $P= 5.00 \times 10^{-5}$
$ E_F = 1.00 eV - (8.62 \times 10^{-5} eV/K)(300K) ln ((5.00 \times 10^{-5})^{-1} - 1)$
$E_F = 0.744 eV$
