Let $ABCDEF$ be a regular hexagon with $AB=1,\ O$ be the center. $\triangle OAB$ is equilateral, radius of the inscribed circle is $\tfrac{\sqrt3}2.$ Area ratio $=\frac{\pi(\frac{\sqrt3}2)^2}{6\times\frac12\times1\times\frac{\sqrt3}2}\approx0.907$

Call the length of one side of the hexagon $x$. Since the apothem of the hexagon is $x\sqrt3$, the area of the circle is $3x^2\pi$. Also, the area of the hexagon is $6x^2\sqrt3$, so the ratio is $\frac{3x^2\pi}{6x^2\sqrt3} = \frac{\pi}{2\sqrt3} > 0.9$, as desired.