jmc

Since $O$ is the center of the circle passing through $A$, $B$, and $C$, $\angle BOC=2\angle BAC=2\cdot 68^{\circ }=136^{\circ }$.



Since $BO=CO$ (both are equal to the circumradius of triangle $ABC$), triangle $BOC$ is isosceles. Hence, $\angle OBC=(180^{\circ }-\angle BOC)/2=(180^{\circ }-136^{\circ })/2=\boxed{22^{\circ }}$.