Parallel circuit with LED on one path and resistor on the other
You can follow jonk's approach and you will end with a Thevenin voltage and a source resistance. 

If you model the LED as an otherwise perfect diode with a 3V drop you will have two possible conditions: 


The Thevenin voltage is less than 3V. In that case you ignore the LED (and, of course, it will be 'off'). 
The Thevenin voltage is greater than 3V. In that case you can calculate the current as the (Vthev - 3V) divided by the Thevenin resistance 




Or you could be lazy and assume the LED is on, meaning the current from the supply is indeed 6V/220 ohms. The current through the parallel resistor is 3V divided by the parallel resistor. The current through the LED is the difference. If that current turns out to be negative, then your initial assumption was wrong. 

The advantage of doing it "properly" is that you can use a graphical load line method, or a nonlinear equation solving method to find the LED current if you model it using the data sheet curves or the Shockley diode equation respectively. 

On the other hand, if it's an exam question I think I'd opt for the cheap and dirty method.