jmc

Triangles WXY and BXY are isosceles triangles that have a leg in common, so they are congruent. Therefore segment $YB$ is equal to a diagonal of square $WXYZ$, so its length is 12 units. By adding point $D$, as shown, we can see that triangles $CDY$ and $YXB$ are similar to triangle $CAB$. This also means that triangle $CDY$ is similar to triangle $YXB$. Since the sides of two similar triangles are related by a constant factor, and we can see that the length of $DY$ is 1/2 the length of $XB$, we know that the length of $CY$ must be $(1/2)(12)=6$ units. Thus, the length of CB is $12+6=\boxed{18\text{ units}}$.