Use zener diode regulation condition

In breakdown region, the zener diode keeps the voltage across itself approximately constant at its zener voltage $V_Z$.

Let the supply voltage be $V$, series resistance be $R$, and load resistance be $R_L$.

Then:

Load current

Since the load is parallel to the zener diode, the voltage across the load is also $V_Z$.

Hence, $I_L=\frac{V_Z}{R_L}$

Zener current

The current through the series resistor splits into load current and zener current: $I=I_L+I_Z$ so, $I_Z=I-I_L$

Substituting the values from the circuit

From the given circuit values, $I=5.8\text{ mA}$ and $I_L=2.5\text{ mA}$

Therefore, $I_Z=5.8-2.5=3.3\text{ mA}$

Match with options

$I_Z=3.3\text{ mA}$

So the correct option is D.