The switch is closed. Compare the brightness of B before and after it is closed.

A) Brighter before

B) No change

C) Brighter after

Solution: A. The current through the battery goes up a bit, but B only gets half of it now. So B gets dimmer. A more rigorous proof is as follows:

Before the switch is closed, the equivalent resistance is 2R (with R being the resistance of a single bulb). So the current (for the battery, A, and B using Ohm's law) will be I = emf/(2R) = (1/2)*emf/R

After the switch is closed, the equivalent resistance is R + R/2 = 3/2*R. The current through the battery will then be I = emf/(3/2*R) = (2/3)*emf/R. B gets half of this current which is (1/3)*emf/R. The current through B decreases, so it gets dimmer when the switch is closed.