The circuit in the figure is that of a resistor R1 in parallel with a set of resistors R2 and R3 that are in series. Across the two there is an applied voltage of 60 V.
The current flowing through R2 and R3 is the same as they are in series. If I is the current flowing through them I*R2 + I*R3 = 60
=> I*R3 + 30 = 60
=> V3 = 60 - 30 = 30
The sum of the voltage drop across the two resistors should be equal to the voltage applied of 60 V across the two. This gives the voltage drop across R3 as 60 - 30 = 30 V.