In the given circuit, the breakdown voltage of the Zener diode is 3...

Given that, ${V}_{z} = 3V$

Let potential at $B = 0V$
Potential at $E\left({V}_{E}\right) = 10V$

Now, ${V}_{C} = {V}_{A} = 3V$ and,

${I}_{z} + {I}_{1} = I . . . \left(1\right)$

The current through $1kΩ$ resistor is given by

$\begin{cases}I & = & \dfrac{10 - 3}{1000}A \\ & = & \dfrac{7}{1000}A\end{cases}$

From the diagram, it follows that

${I}_{1} = \dfrac{3}{2000}A$

Therefore, from equation (1), it follows that

$\begin{cases}{I}_{z} & = & \dfrac{7 - 1 . 5}{1000}A \\ & = & 5 . 5mA\end{cases}$