Answer
$I = 5 A$
This current flows toward the junction, that is, upward
Work Step by Step
The current in the $2\text{ }\Omega$ resistor is $I_1 = 4\text{ V}/2\text{ }\Omega = 2 \text{A} $ to the left. The current in the $5\text{ }\Omega$ is $ I_2 = 15\text{ V}/5\text{ }\Omega = 3 \text{A} $. Using Kirchoff's juction law, we see that $$I = I_1 + I_2 = 2 \text{A} + 3 \text{A} = 5 \text{A}$$ This current flows away from the junction, to the left.
