Use Zener breakdown condition

Since the Zener diode has reverse breakdown voltage $V_Z = 5.6\,\text{V}$ in breakdown, the voltage across the Zener remains approximately $5.6\,\text{V}$.

Find the voltage across the series resistor

From the given circuit, the supply is $10\,\text{V}$ and the series resistor is $220\,\Omega$.

Hence voltage across the resistor is $V_R = 10 - 5.6 = 4.4\,\text{V}$

Find the current through the resistor

By Ohm’s law, $I_R = \frac{V_R}{R} = \frac{4.4}{220} = 0.02\,\text{A} = 20\,\text{mA}$

Find load current

From the circuit, the load resistor is $560\,\Omega$ and it is in parallel with the Zener, so it also has $5.6\,\text{V}$ across it.

Therefore, $I_L = \frac{5.6}{560} = 0.01\,\text{A} = 10\,\text{mA}$

Find Zener current

The series current splits into load current and Zener current: $I_R = I_L + I_Z$

So, $I_Z = I_R - I_L = 20 - 10 = 10\,\text{mA}$

Match with options

$I_Z = 10\,\text{mA}$

Therefore the correct option is D.