In the figure, the reverse breakdown voltage of a Zener diode is 5.6 V, then the current $I_z$ through the diode is

The potential difference across the Zener diode is 5.6V.

Since the resistor is connected in parallel combination.

$I_L=\frac{V_Z}{R_L}=\frac{5.6}{800}\mathrm{A}=7\mathrm{mA}$

the current flowing through the $200\Omega$

$I=\frac{V-V_z}{R_s}=\frac{9-5.6}{200}A=17mA$

$\Rightarrow I_z=(17-7)mA=10mA$