(A) Given: $N_{Si} = 5 \times 10^{28} \text{ atoms/m}^3$,$n_i = 1.5 \times 10^{16} \text{ m}^{-3}$.
$(i)$ For $n$-type region:
Concentration of $As$ (donor) atoms $N_D = 1 \text{ ppm} = 10^{-6} \times 5 \times 10^{28} = 5 \times 10^{22} \text{ atoms/m}^3$.
Electron density $n_e \approx N_D = 5 \times 10^{22} \text{ m}^{-3}$.
Hole density $n_h = n_i^2 / n_e = (1.5 \times 10^{16})^2 / (5 \times 10^{22}) = 4.5 \times 10^9 \text{ m}^{-3}$.
For $p$-type region:
Concentration of $B$ (acceptor) atoms $N_A = 200 \text{ ppm} = 200 \times 10^{-6} \times 5 \times 10^{28} = 10^{25} \text{ atoms/m}^3$.
Hole density $n_h \approx N_A = 10^{25} \text{ m}^{-3}$.
Electron density $n_e = n_i^2 / n_h = (1.5 \times 10^{16})^2 / 10^{25} = 2.25 \times 10^7 \text{ m}^{-3}$.
$(ii)$ In reverse bias,the reverse saturation current is primarily due to the flow of minority charge carriers (electrons in the $p$-region and holes in the $n$-region) across the junction.