Step 1: Use shape to judge polarity.
SO$_2$ is bent (V-shaped); its bond dipoles do not cancel $\Rightarrow$ molecule is polar.
Step 2: Eliminate others.
SO$_3$ (trigonal planar), BF$_3$ (trigonal planar) and CO$_2$ (linear) have symmetric geometries $\Rightarrow$ overall non-polar.
Step 3: Conclusion.
Therefore, the polar compound here is SO$_2$.