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 opposing each other. so the direction of the current in this circuit loop is the direction of the current from the big battery voltage.
Hence, the direction of the current in this circuit loop is counterclockwise.
Therefore, the direction of the current in the resistors is toward the left.
From Kirchhoff's loop law, starting from the right lower corner and moving counterclockwise.
$$\Delta V_{\rm loop}=18-IR-9=0$$
Hence,
$$I=\dfrac{9}{R}=\dfrac{9}{10}=\color{red}{\bf 0.9}\;\rm A$$
$$\color{blue}{\bf [b]}$$
The author assumed that the left lower corner is the 0-V.
