Answer:
idk b idk ssry dude good luck tho
The work done by the battery is equal to the charge transferred during the process times the potential difference between the two terminals of the battery:

where q is the charge and

is the potential difference.
In our problem, the work done is W=39 J while the potential difference of the battery is

, so we can find the charge transferred by the battery:
1 ampere = 1 coulomb of charge per second
0.120 ampere = 0.120 coulomb of charge per second
(0.120 C/sec) x (900 sec) = <em>108 Coulombs .</em>
The voltage of the battery, generator, or leiden jar
that's supplying the current is irrelevant.