Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
It is obvious, from the given figure, that the two batteries are in series, so the direction of the current in this circuit loop is clockwise.
So the direction of the current in the resistors is toward the right.
Its magnitude is given by Ohm's law,
$$\Delta V_{net}=IR$$
$$I=\dfrac{\Delta V_{net}}{R}$$
Plug the known;
$$I=\dfrac{6+3}{18}=\color{red}{\bf 0.5}\;\rm A$$
This circuit is too simple to use Kirchhoff's loop law, but you can indeed use it.
$$\color{blue}{\bf [b]}$$
The author assumed that the left lower corner is the 0-V.