The voltmeter is ideal so it doesn't draw any current from the circuit.
Potential difference across 9 ft resistance (given)
So, the current in the resistance
i.e.,
The same current passes through . The resistors and are in series and their equivalent, i.e., is in parallel with between A and B . Dividing the current in the inverse ratio of resistances between A and B .

And
at junction divides into three parts. Since the resistances are in parallel between C and D, current will distribute in the inverse ratio of resistances.

So
( k being a constant of proportionality)
and
but

or
so
Thus current through resistance is 0.5 A