For the given circuit the power dissipated by Zener diode is

Current flowing through load resistor is given as $\mathrm{i}=\frac{16}{1000+1200}=7.2\mathrm{mA}$

Voltage drop across the load resistor, ${\mathrm{V}}_{\mathrm{ab}}={\mathrm{iR}}_{\mathrm{L}}=7.2\times {10}^{-3}\times 1200=8.72 \mathrm{V}$

$\text{ As }{\mathrm{V}}_{\mathrm{ab}}<10\mathrm{V}$

So, breakdown of zener diode does not occur and no current flows through it.

$\text{ So, }{\mathrm{I}}_{\mathrm{z}}=0\text{ and }{\mathrm{P}}_{\mathrm{z}}={\mathrm{I}}_{\mathrm{Z}}{\mathrm{V}}_{\mathrm{z}}=0\mathrm{W}$