Denote the lengths of the sides of the inscribed deltoid by $a$ and $2a$. Due to symmetry, the longer of the two diagonals passes through the centre of the circle. By Thales' theorem there is a right angle between the sides of length $a$ and $2a$. By Pythagoras' theorem the length of the longer diagonal is equal to $\sqrt{a^2 + (2a)^2} = \sqrt{5a^2} = a\sqrt{5}$, which implies that $r = \frac{a\sqrt{5}}{2}$. The area of the deltoid is equal to $2 \cdot \frac{a \cdot 2a}{2} = 2a^2$ and the area of the circle is $\pi r^2 = \frac{5\pi a^2}{4}$. The ratio is $\frac{8}{5\pi}$.