Can someone please explain Example 14.5 of Semiconductor Electronic of Class 12 NCERT Physics Book part 2 on page 486. I am not able do understand the solution.

Here, GIven Values are, 

Load Current is I_L=4.0*10^-3A, Zener Voltage is V_Z=6.0V, unregulated Input Voltage is , V_U=10.0V,

Since Zener Voltage remains constant throughout the Voltage RAnge, Hence, Voltage dross across Series Resistor V_s=4.0V

Now, the Series Resistor, must have a greater value than Zener Diode so as to have a better Load COntrol.

Let us assume Current in Resistor to be 5x the current in Zener Diode, i.e.I_s=5*4mA or 20*10^-3 A

Hence, Resistance of Series resistor= Vs/Is= 4/24*10^-3= 166.67 ohm.

WE can approximate  the value of given resistor to be 150 ohm which would be our required answer.