jmc

Let $O$ be the center of the circle and $A,B$ be two points on the circle such that $\angle AOB=\theta$. If the circle circumscribes the sector, then the circle must circumscribe $△AOB$. Draw the perpendicular bisectors of $OA$ and $OB$ and mark the intersection as point $C$, and draw a line from $C$ to $O$. By HL Congruency and CPCTC, $\angle AOC=\angle BOC=\theta /2$. Let $R$ be the circumradius of the triangle. Using the definition of cosine for right triangles,$$\cos (\theta /2)=\frac{3}{R}$$$$R=\frac{3}{\cos (\theta /2)}$$$$R=3\sec (\theta /2)$$Answer choices A, C, and E are smaller, so they are eliminated. However, as $\theta$ aproaches $90^{\circ }$, the value $3\sec \theta$ would approach infinity while $3\sec \frac{1}{2}\theta$ would approach $\frac{3\sqrt{2}}{2}$. A super large circle would definitely not be a circumcircle if $\theta$ is close to $90^{\circ }$, so we can confirm that the answer is $\boxed{3\sec \frac{1}{2}\theta }$.