Given that $n_i = 1.5 \times 10^{16} \;/m^3$, the thermally generated electrons are negligibly small compared to those produced by doping. Therefore, $n_e \approx N_D$

Since $n_e\;n_h=n_i^2$, the number of holes $n_h = \large \frac {2.25 \times 10^{32}}{5 \times 10^{22}}$$ \approx 9 \times 10^9 \;m^{-3}$