If you are only permitted to use a full-wave rectifier, and a zener diode...
You are dropping a lot of volts, so there needs to be a robust volt-dropping resistor.
Each diode in the full-wave rectifier needs to withstand the peaks of 240 VAC. This is 340V.
When the supply is minimum (100 VAC), peaks will be 140V.
Although the zener diode may be rated 11 V for low current levels, it may also carry high current which may elevate its operating voltage higher than that.
The load calculates to be 20 ohms (11/.55).
A safety resistor is customary above the zener diode. Whatever its value, it must drop as much as 329 V while carrying more than 550 mA. Its rating will end up needing to be on the order of a few hundred watts (even though it will not be carrying that entire amount every moment).
You want the zener diode to carry some amount of current at all expected supply voltages. Maybe 25 mA minimum (5% of the load current) when the supply is 100 VAC (peaks at 140VDC). It will need to be rated 305 mW minimum.
When the supply is the maximum expected (240 VAC), a whole new round of calculations is needed.