A Zener diode of breakdown voltage 10V is used as a voltage...

As Zener is in breakdown region, therefore potential drop across it would be $10V$. Now, 

${I}_{3} = \dfrac{10}{500} = \dfrac{1}{50}A$ and 

${I}_{1} = \dfrac{10}{200} = \dfrac{1}{20}A$

Also, ${I}_{2} = {I}_{1} - {I}_{3}$

$⇒ {I}_{2} = \left(\dfrac{1}{20} - \dfrac{1}{50}\right) = \left(\dfrac{3}{100}\right)A = 30mA$