Solution:
The angle measure is $30^{\circ}$.

Let $O$ be the circumcenter of the triangle $A B M$. From $\angle B A M=105^{\circ}$ follows $\angle M B O=15^{\circ}$. Let $M', C'$ be the projections of points $M, C$ onto the line $B O$. Since $\angle M B O=15^{\circ}$, then $\angle M O M'=30^{\circ}$ and consequently $M M'=\frac{M O}{2}$. On the other hand, $M M'$ joins the midpoints of two sides of the triangle $B C C'$, which implies $C C'=M O=A O$.
The relation $\angle M A C=\angle A B C$ implies $C A$ tangent to $\omega$, hence $A O \perp A C$. It follows that $\triangle A C O \equiv \triangle O C C'$, and furthermore $O B \parallel A C$.
Therefore $\angle A O M=\angle A O M'-\angle M O M'=90^{\circ}-30^{\circ}=60^{\circ}$ and $\angle A B M=\frac{\angle A O M}{2}=30^{\circ}$.