Answer
The current in the wire to the right of the junction is $1~A$ and this current flows to the left (into the junction).
Work Step by Step
We can find the current across the $2~\Omega$ resistor:
$I = \frac{V}{R} = \frac{4~V}{2~\Omega} = 2~A$
Since the current flows from higher potential to lower potential, this current flows to the right (toward the junction).
We can find the current across the $5~\Omega$ resistor:
$I = \frac{V}{R} = \frac{15~V}{5~\Omega} = 3~A$
Since the current flows from higher potential to lower potential, this current flows downward (away from the junction).
The total current flowing into a junction must be equal to the total current flowing out of the junction. Since the current flowing into the junction must be $3~A$, the current in the wire to the right of the junction is $1~A$ and this current flows into the junction.
