Current electricity

The crucial point lies on the statement:
"the light it (the replaced bulb) gives is very dim, although the other bulbs light up brightly"

Since "the other bulbs light up brightly", statements (1) and (2) are clearly not correct.

Thus, only statement (3) is right
If the repaced bulb is designed to run at voltage lower than 10 v, in order to give a power of 5 w as before, the resistance of the bulb will be lower (power = voltage^2/resistance)

A bulb of lower resistance than the other 19 bulbs in a series circuit will share a voltage less than 10 v. Because of the introduction of a lower resistance bulb into the circuit, the toal resistance of the circuit is lower than before, the current is thus higher than before. This is the why the other 19 bulbs are brighter.