A network of four resistances is connected to 9V battery, as shown...

The resistance in each branch is $6Ω$. They are in parallel. So, net resistance is $R ' = \dfrac{6}{2} = 3Ω$.

The total current flowing through the loop,

$I = \dfrac{V}{R '} = \dfrac{9V}{3Ω} = 3A$

Thus, through each branch the current is $1 . 5A$.

Hence, the voltage difference across the points is

${V}_{A} - {V}_{B} = \left(4 \times 1 . 5\right) - \left(2 \times 1 . 5\right) = 3V$