(N/A) Since the air is less dense at higher altitudes,the pressure is also lower.
$(a)$ Consider a horizontal portion of air with cross-sectional area $A$ and height $dh$. Let the pressure on the top surface be $P$ and on the bottom surface be $P + dP$. If the portion is in equilibrium,the net upward force must be balanced by the weight.
$(P + dP)A - PA = -mg$ (where mass = volume $\times$ density)
$(dP)A = -\rho(A dh)g$
$dp = -\rho g dh$ ... $(1)$
The negative sign indicates that pressure decreases as height increases.
$(b)$ Let the density of air on the earth's surface be $\rho_0$. Given $P \propto \rho$,we have $\frac{P}{P_0} = \frac{\rho}{\rho_0}$,so $\rho = \left(\frac{P}{P_0}\right)\rho_0$ ... $(2)$
Substituting $(2)$ into $(1)$:
$dP = -\left(\frac{P}{P_0}\right)\rho_0 g dh$
$\frac{dP}{P} = -\frac{\rho_0 g}{P_0} dh$
Integrating both sides from $0$ to $h$:
$\int_{P_0}^{P} \frac{dP}{P} = -\frac{\rho_0 g}{P_0} \int_{0}^{h} dh$
$\ln\left(\frac{P}{P_0}\right) = -\frac{\rho_0 g h}{P_0}$
$P = P_0 e^{-\frac{\rho_0 g h}{P_0}}$
$(c)$ Given $P = \frac{P_0}{10}$,$\ln\left(\frac{1}{10}\right) = -\frac{\rho_0 g h}{P_0}$
$h = \frac{P_0 \ln(10)}{\rho_0 g} = \frac{1.03 \times 10^5 \times 2.303}{1.29 \times 9.8} \approx 18750 \text{ m} \approx 18.75 \text{ km}$.
$(d)$ The assumption is that the density of air is proportional to pressure,which implies an isothermal atmosphere (constant temperature),whereas the actual temperature of the atmosphere varies with height.