Correct Answer - B
Here, `R_L=5xx10^3 Omega , V_i=220 V`,
Zener voltage ,`V_Z=50 V`
Load current, `I_L=V_Z/R_L=50/(20xx10^3)=2.5xx10^(-3) A`
Current through R, `I=(220-50)/(5xx10^3)=34xx10^(-3) A`
Zener current,
`I_Z=I-I_L=34xx10^(-3)-2.5xx10^(-3)`
`=31.5xx10^(-3)A =31.5 mA`