Zener diode in breakdown

Since the Zener breakdown voltage is given as $V_Z = 3.0\text{ V},$ when it is operating in breakdown, the voltage across it remains approximately $3.0\,$V.

Apply KVL to the circuit

Let the supply voltage be $V$ and the series resistor be $R$ as shown in the circuit. Then the current through the resistor is $I = \frac{V - V_Z}{R}.$

Load current

If a load resistor is connected in parallel with the Zener, then the load current is $I_L = \frac{V_Z}{R_L}.$

Zener current

The series current splits as $I = I_Z + I_L.$ Hence, $I_Z = I - I_L.$

Using the circuit values

From the given circuit values, $I = 10\text{ mA}, \qquad I_L = 4.5\text{ mA}.$ Therefore, $I_Z = 10 - 4.5 = 5.5\text{ mA}.$

Final answer

$\boxed{I_Z = 5.5\text{ mA}}$

So the correct option is C.