When zener diode works in breakdown state, voltage across the zener will remain same.

\(\therefore \) \({V_{{\text{across}}}}4{\mkern 1mu} K\Omega  = 6{\text{V}}\)

\(\therefore \) Current through \(4\, K\Omega = \frac{6}{4000}\,A\) \(=\frac{6}{4}\,mA\)

since input voltage \(=16\,V\)

\(\therefore \) Potential difference across \(2 \mathrm{K} \Omega=10 \mathrm{V}\)

\(\therefore \) Current through \(2 \mathrm{k} \Omega=\frac{10}{2000}=5 \mathrm{mA}\)

\(\therefore \) Current through zener diode \(=\left(I_{s}-\mathrm{I}_{\mathrm{L}}\right)=3.5 \mathrm{mA}\)