Answer
Bulb $B_1$ increases in brightness after the switch is closed.
Work Step by Step
When the switch is closed, bulb $B_2$ is effectively removed from the circuit because the switch acts as a short circuit. That is, the current would rather pass through the zero-resistance switch than through bulb $B_2$.
This lowers the equivalent resistance of the circuit, so the current that the battery supplies will increase.
All of the current passes through bulb $B_1$, so it increases in brightness after the switch is closed.
