Maximum current will flow from zener if input voltage is maximum.

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 $=\left(I_{s}-\mathrm{I}_{\mathrm{L}}\right)=3.5 \mathrm{mA}$