Silicon has 5.0 × 10²⁸ atoms in each cubic metre…

Step 1. The holes. $N_A = \dfrac{5.0 \times 10^{28}\ \text{m}^{-3}}{2.0 \times 10^7} = 2.5 \times 10^{21}\ \text{m}^{-3}$, and $n_h \approx N_A$.

Step 2. The electrons. $n_e = \dfrac{n_i^2}{n_h} = \dfrac{(1.0 \times 10^{16}\ \text{m}^{-3})^2}{2.5 \times 10^{21}\ \text{m}^{-3}} = 4.0 \times 10^{10}\ \text{m}^{-3}$.

Step 3. The conductivity. $\sigma = e\,(n_e \mu_e + n_h \mu_h)$. The hole term is $1.602 \times 10^{-19}\ \text{C} \times 2.5 \times 10^{21}\ \text{m}^{-3} \times 0.048\ \text{m}^2/\text{V s} = 19.2$ S/m. The electron term is about $9 \times 10^{-10}$ S/m and can be left out. So $\sigma = 19.2$ S/m.

Step 4. The ratio. $\dfrac{19.2\ \text{S/m}}{2.93 \times 10^{-4}\ \text{S/m}} = 6.56 \times 10^4$.