Answer
$I = 3~A$
Work Step by Step
From ground to a point to the left of the $2~\Omega$ resistor, there is a potential increase of 9 V. Therefore, the potential at that point is 9 V.
From ground to a point to the right of the $2~\Omega$ resistor, there is a potential increase of 3 V. Therefore, the potential at that point is 3 V.
The potential difference across the $2~\Omega$ resistor is $6~V$. We can find the current:
$I = \frac{V}{R} = \frac{6~V}{2~\Omega} = 3~A$
