Packed in the middle

Let $r$ be the radius of the green inner circle and $d$ the distance between $A$ and the inner circle's center.

Let $A_{blue}$ be the area of the circular sector, $A_{red}$ the area of the triangle $\triangle {ABC}$ and $A_{green}$ the area of the inner circle.

We can write $\sin(\angle{CAB}) = \sin(\frac{\pi}{6}) = \frac{r}{d} \Rightarrow d = 2r$ .

Since $A_{blue} = \frac{\pi (d+r)^2}{6} = \frac{A_{red}}{2}$ , then $A_{red} = \frac{\pi (d+r)^2}{3} = \frac{\pi (3r)^2}{3} = 3\pi r^2 = 3*A_{green} \Rightarrow \frac{A_{red}}{A_{green}}= \boxed{3}$