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The formula you are looking for is W = E(squared)/R, W = Voltage (squared) divided by Resistance.
Just use Ohm's Law Voltage = Current x Resistance Amps = Voltage Divided By Resistance Amps = 120 / 260
Let R be the resistance, V the voltage, and I the current R = V / I R = 120 / 24 R = 5 ohm
V = I x R so current I = 1/2 amp. I bet the bulb is rated at 60 W because Watts = Current x Voltage. Where V = voltage (volt) I = current (ampere) R = resistance (ohm) Your question isn't easy to answer. A lamp has two 'resistances': a 'cold' resistance, and a 'hot' resistance. Before it is energised, it is cold, so its resistance is low; when it is energised, it becomes very hot, and its resistance increases significantly. So, the question is whether your '240 ohms' is the cold resistance or the hot resistance. If it is the cold resistance, then a current of 0.5 A will flow through it for a fraction of a second, then its resistance will increase significantly, and the current will fall to a very much smaller value.
Just get out your trusty calculator and multiply 120 times 18.