jmc

Label the lower right corner of the square point $D$ and the lower left corner $E$. The interior angles of a regular hexagon are 120 degrees and the interior angles of a square are 90 degrees. Thus, $m\angle BDC=m\angle BDE-m\angle CDE=120^{\circ }-90^{\circ }=30^{\circ }$. In addition, because the square and regular hexagon share a side and all of their sides have the same length, segments $CD$ and $BD$ have the same length. Thus, triangle $BCD$ is isosceles.

Because the base angles of an isosceles triangle are congruent, $m\angle BCD=m\angle CBD=x$. In addition, because the interior angles of a triangle sum to 180 degrees, we have $$\begin{array}{rl}180^{\circ }& =m\angle BDC+m\angle BCD+m\angle CBD\\ & =30^{\circ }+x+x\\ 150^{\circ }& =2x\\ 75^{\circ }=x& \end{array}$$ Thus, $m\angle CBD=75^{\circ }$. Finally, we calculate $m\angle ABC=m\angle ABD-m\angle CBD=120^{\circ }-75^{\circ }={\boxed{45}}^{\circ }$.