A silicon wafer is doped with both kinds of impurity…

Step 1. The type and the electrons. $N_D > N_A$, so the wafer is n-type. $n_e \approx N_D - N_A = 2.0 \times 10^{21}\ \text{m}^{-3}$.

Step 2. The holes. $n_h = \dfrac{n_i^2}{n_e} = \dfrac{(1.0 \times 10^{16}\ \text{m}^{-3})^2}{2.0 \times 10^{21}\ \text{m}^{-3}} = 5.0 \times 10^{10}\ \text{m}^{-3}$.

Step 3. The conductivity. $\sigma \approx e\, n_e \mu_e = 1.602 \times 10^{-19}\ \text{C} \times 2.0 \times 10^{21}\ \text{m}^{-3} \times 0.135\ \text{m}^2/\text{V s} = 43.3$ S/m.

Step 4. The resistivity. $\rho = \dfrac{1}{\sigma} = \dfrac{1}{43.3\ \text{S/m}} = 0.0231\ \Omega\ \text{m} = 2.31\ \Omega\ \text{cm}$.

Step 5. After more acceptors. Now $N_A = 5.0 \times 10^{21}\ \text{m}^{-3}$ is larger than $N_D$. The wafer is p-type, with $n_h \approx 5.0 \times 10^{21} - 3.0 \times 10^{21} = 2.0 \times 10^{21}\ \text{m}^{-3}$. So $\sigma \approx e\, n_h \mu_h = 1.602 \times 10^{-19}\ \text{C} \times 2.0 \times 10^{21}\ \text{m}^{-3} \times 0.048\ \text{m}^2/\text{V s} = 15.4$ S/m.