Answer: 4.

Solution. Let the radius of the small semicircle be $x$, the center of the large semicircle be $A$, the center of the circle be $B$, and the center of the small semicircle be $C$. The centers of the tangent circles and semicircle, and the corresponding points of tangency, lie on the same straight line, so $A B C$ is a right triangle with legs $A B=6, A C=12-x, B C=6+x$. By the Pythagorean theorem, we have $(x+6)^{2}=6^{2}+(12-x)^{2}$, from which $x=4$.

Criteria. It should be clearly explained why $A B=6, A C=12-x, B C=6+x$, otherwise 2 points will be deducted.
9.3. Inside a semicircle of radius 12, there are a circle of radius 6 and a small semicircle, each touching the others pairwise, as shown in the figure. Find the radius of the small semicircle.