Answer
The bulb becomes brighter.
Work Step by Step
When the switch is closed, the equivalent resistance of the circuit decreases. Previously, two identical resistances were in parallel with each other, and the combination was in series with another resistance and a bulb ($R_{eq}=\frac{R}{2}+R+R_{bulb}$), and now there are 3 in parallel ($R_{eq}=\frac{R}{3}+R+R_{bulb}$).
The resistance decreases, so the current in the circuit increases. The bulb becomes brighter.
The answer is the same even if the resistances aren't identical; adding another resistance in parallel always decreases the resistance of any parallel combination.