When fully discharged, an accumulator battery of emf 6 V possesses...

The emf of the battery is $E = 6 \text{ V}$.

The constant potential difference provided by the charger is $V = 9 \text{ V}$.

When the battery is completely charged, its internal resistance drops to $r = 1 \Omega$.

During the charging process, the current $I$ flows into the positive terminal of the battery. The net potential difference driving the current is $V - E$.

Using Ohm's law, the current passing through the battery is given by:

$I = \dfrac{V - E}{r}$

Substituting the given values:

$I = \dfrac{9 - 6}{1} = 3 \text{ A}$