I'll try to explain:
The 12W lamp needs a certain current, as well as the 60W lamp. If you calculate the currents the lamps need, you should find a difference. So where to put the excess power? Now the 2 Ohm resistor comes into play, it diverts the excess current and that is its task in this circuit.
Understanding this requires Kirchoff's and Ohm's laws:
Kirchhoff's current law : The sum of all currents in a node is equal to zero. This means that all currents flowing into a node flow out again. So in your circuit the current coming from the 60W lamp has to be divided between the 2 Ohm resistor and the 12W lamp.
Kirchhoff's voltage law: The sum of all voltages in a mesh is equal to zero. Applied to your circuit it means, that the 24 V supply voltage must be distributed to all components. e.g. the 12V at the 60W lamp + the 6V at the 12W lamp + the voltage at the 1.2 Ohm resistor must be equal to the 24 V
Furthermore you have to know that in a parallel circuit all voltages are the same. Applied to your circuit this means that the 2 Ohm resistor and the 12W lamp have the same voltage.
Furthermore, in a series circuit the current is the constant through all the elements, in your circuit the 1.2 Ohm resistor and the 60W lamp are passed through by the same current.
In addition you need to know that the voltage is directly proportional to resistance, which means that the largest resistor will drop the largest voltage.
Power is equal to current times voltage W = V*I
Ohm's law: I = V/R
I think that should be enough to solve the problem.