You've done all the hard work and now there's one little part left.
You're being asked, given a certain battery capacity in amp hours and a certain current in amps flowing out of the battery, for how many hours (or seconds) will the current flow? The assumption is that as long as the current is flowing the light will be on and as soon as the capacity of the battery is drained the light will go off.
You know the power the bulb will use. You know the voltage of the battery, you know the total charge store in the battery in the form "amp hours" (A hr). Amp hours are an engineering term for charge and are equivalent to coulombs.
1 amp flowing for 1 hour = 1 coulomb / second flowing for 3600 seconds = 3600 C
A battery capacity of 1 amp hour means that:
- a current of 1 amp can flow for 1 hour
- a current of 2 amps can flow for 0.5 hours
- a current of 0.25 amps can flow for 4 hours
Notice that in each of these the current (amps) * time (hours) = capacity (amp hours): 1 * 1 = 2 * 0.5 = 0.25 * 4 = 1
A constant potential difference across the light means that the current will also be constant. In the real world as batteries discharge the voltage across their terminals decreases. This drop in voltage is due to internal reistance of the battery increasing as chemical reactions take place inside the battery but that's outside the scope of this question. If the driving voltage wasn't contant the current would also change which makes the problem harder.