Reason to explain why $\mathit{n}-\mathit{p}$ region of zener diode is heavily doped
Calculation of current through zener diode $\mathit{n}$ and $\mathit{p}$ regions of zener diode are heavily doped so that depletion region formed is very thin and electric field at the junction is extremely high even for a small reverse bias voltage.
Current in the circuit is :
$\mathit{I}=\frac{\mathit{V}}{\mathit{R}}=\frac{5}{250}=\frac{1}{50}$ $=0.02\mathrm{A}$
Current through resistor of $1\mathrm{k}\mathit{\Omega }$ is:
$\mathit{I}=\frac{15}{1000}=0.015\mathrm{A}$
As zener diode and $1\mathrm{k}\mathit{\Omega }$ resistor are in parallel, current through the zener diode is :
$\mathit{I}=0.02-0.015=0.005\mathrm{A}$