The current through \(1 \mathrm{k} \Omega\) is

\(I^{\prime}=\frac{15 \mathrm{V}}{1 \times 10^{3} \Omega}=15 \times 10^{-3} \mathrm{A}=15 \mathrm{mA}\)

The voltage drop across \(250 \Omega=20 \mathrm{V}-15 \mathrm{V}=5 \mathrm{V}\)

The current through \(250 \Omega\) is

\(I=\frac{5 \mathrm{V}}{250 \Omega}=0.02 \mathrm{A}=20 \mathrm{mA}\)

The current through the zener diode is

\(I_{2}=I-I^{\prime}=(20-15) \mathrm{mA}=5 \mathrm{mA}\)