Given, zener breakdown voltage of zener diode, $V_z=5 \mathrm{V}$.

Maximum power dissipation by the zener diode, $P_{\max }=240 \mathrm{mW}$.

Let the maximum current the diode can handle be $I_{\max }$.

The power dissipated in the diode is given as $P_{\max }=V_z{I}_{\max }\Rightarrow I_{\max }=\frac{P_{\max }}{V_z}=\frac{240}{5}=48 \mathrm{mA}$