In the figure shown, the currents through the series resistance and load respectively

(1) 9 mA, 14 mA                 (2) 14 mA, 5 mA               (3) 1 mA, 14 mA               (4) 1 mA, 6 mA

Dear Student 

Here zener diode is in reverse bias so will work in breakdown state as voltage regulator . Hence voltage across load is 50 V 
using V=IR_L 
50 = I x 10000
I = 0.005A = 5 mA 

Now Voltage across series resistance is 120 - 50(zener voltage) = 70 V 
so V_s = I_sR_s 
     70 = I_s x 5000
      I_s = 70/5000
          = 14 mA 
So ans is (2) 14 mA, 5 mA

Regards