Extend the frustum to a full cone (apex slant distances $6$ and $12$); it unrolls to a sector of angle $\pi$

The straight line to the honey cuts across the missing top cone, so the path runs tangent to the inner circle of radius $6,$ then along its arc

Extend the frustum to a full cone. Since the radii are $3$ and $6$ with slant band $6,$ the apex is slant distance $6$ from the top rim and $12$ from the bottom rim. The bottom circumference $12\pi$ unrolls to a sector of radius $12$ and angle $\dfrac{12\pi}{12}=\pi.$ Place the bug at $(12,0)$ in this pattern; the honey, halfway around the base, is at radius $6$ and angle $\tfrac{\pi}{2}.$ The straight chord between them passes within radius $6$ (off the surface), so the geodesic goes tangent to the circle of radius $6:$ the tangent has length $\sqrt{12^2-6^2}=6\sqrt3$ and touches at angle $\tfrac{\pi}{3},$ after which the path follows the arc of angle $\tfrac{\pi}{6}$ on radius $6,$ of length $6\cdot\tfrac{\pi}{6}=\pi.$ The shortest path is $6\sqrt3+\pi.$

Thus, the correct answer is E.