Ohm's law allows us to find the equivalent resistance of the circuit. In fact:

where
V is the voltage of the battery
I is the current in the circuit

is the equivalent resistance
Re-arranging the equation and using the data of the problem, we get

The three resistors are in series, so their equivalent resistance is just the sum of the three resistances:

Since we know the value of the equivalent resistance,

, we can find the value of R3:
