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

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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 $M^{\prime}, C^{\prime}$ be the projections of points $M, C$ onto the line $B O$. Since $\angle M B O=15^{\circ}$, then $\angle M O M^{\prime}=30^{\circ}$ and consequently $M M^{\prime}=\frac{M O}{2}$. On the other hand, $M M^{\prime}$ joins the midpoints of two sides of the triangle $B C C^{\prime}$, which implies $C C^{\prime}=M O=A O$.

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 $\triangle A C O \equiv \triangle O C C^{\prime}$, and furthermore $O B \| A C$.

Therefore $\angle A O M=\angle A O M^{\prime}-\angle M O M^{\prime}=90^{\circ}-30^{\circ}=60^{\circ}$ and $\angle A B M=$ $\frac{\angle A O M}{2}=30^{\circ}$.