Capacitor

In given circuit capacitor does not conducts once it is charged. So in steady state we have following circuit. ⇒ Now, let I= current drawn from cell. Then we have above current distribution.
Using KVL in loop ABCDEFGA; we have;
‌−6I−1‌

I

2

−‌

3I

2

−I+9=0
⇒‌‌9I=9‌ or ‌I=1A
Now, potential difference between B and D (using KVL);
‌VB−6×1−1×‌

1

2

=VD
⇒VB−VD=6+‌

1

2

=‌

13

2

=6.5V