$\text{From the figure the P.D across }2k\Omega \text{ resistor is 10V>}$

$I_L = \frac{V_z}{R_L} = \frac{10V}{2k\Omega} = 5mA$

$\text{P.D. across }4 K\Omega \text{ is 50V}$

$\Rightarrow I = \frac{50V}{4000\Omega} = 12.5mA$

$\Rightarrow I_Z = I - I_L = 7.5mA$