Silicon has 5.0 × 10²⁸ atoms in each cubic metre…

Step 1. The donors. $N_D = \dfrac{5.0 \times 10^{28}\ \text{m}^{-3}}{2.0 \times 10^8} = 2.5 \times 10^{20}\ \text{m}^{-3}$.

Step 2. The electrons. Each donor gives one electron, and $N_D \gg n_i$. So $n_e \approx 2.5 \times 10^{20}\ \text{m}^{-3}$.

Step 3. The holes. $n_h = \dfrac{n_i^2}{n_e} = \dfrac{(1.0 \times 10^{16}\ \text{m}^{-3})^2}{2.5 \times 10^{20}\ \text{m}^{-3}} = 4.0 \times 10^{11}\ \text{m}^{-3}$.