Answer: D

Small Hint:

Both arcs have radius $AB,$ and $C$ is at distance $AB$ from both $A$ and $B,$ so $\triangle ABC$ is equilateral

Big Hint:

Use Power of a Point at $A$ with $AD = \tfrac12 AB$ to find the small circle’s radius $r = \tfrac38 AB$

Solution:

Each arc has radius $AB,$ and $C$ is at distance $AB$ from both $A$ and $B,$ so $\triangle ABC$ is equilateral. Thus arc $BC$ subtends $60^\circ$ of a circle of radius $AB,$ whose full circumference is $6 \cdot 12 = 72.$

Let the small circle have radius $r$ and touch $\overline{AB}$ at its midpoint $D,$ where $AD = \tfrac12 AB.$ By Power of a Point, $AD^2 = AB(AB - 2r),$ so $$\frac{AB^2}{4} = AB^2 - 2r\,AB,$$ giving $2r = \tfrac34 AB,$ hence $r = \tfrac38 AB.$

The circumferences are in the ratio of the radii, so the small circle’s circumference is $\tfrac38 \cdot 72 = 27.$

Thus, the correct answer is D.