The brightness of the lamp is proportional to the current flowing through the lamp: the larger the current, the brighter the lamp.
The current flowing through the lamp is given by Ohm's law:
[tex]I=\frac{V}{R}[/tex]
where
V is the potential difference across the lamp, which is equal to the emf of the battery, and R is the resistance of the lamp.
The problem says that the battery is replaced with one with lower emf. Looking at the formula, this means that V decreases: if we want to keep the same brightness, we need to keep I constant, therefore we need to decrease R, the resistance of the lamp.
