smc

$△OCD$ is copied six times to form the hexagon, so if we find the ratio of the area of the kite inside $△OCD$ to the the area of $△OCD$ itself, it will be the same ratio. Let $OC=CD=OD=2$ so that the area of the triangle is $\frac{\sqrt{3}{s}^2}{4}=\sqrt{3}$. Notice that $△OCD$ is made up of a kite and two $30-60-90$ triangles. The two hypotenuses of these two triangles form $CD$, so the hypotenuse of each triangle must be $\frac{2}{2}=1$. Thus, the legs of each triangle are $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$, and the area of two of these triangles is $\frac{1}{2}\cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}$. Subtracting the area of the two triangles from the area of the equilateral triangle, we find that the area of the kite is $\sqrt{3}-\frac{\sqrt{3}}{4}=\frac{3\sqrt{3}}{4}$. Thus, the ratio of areas is $\frac{3}{4}$, which is option $\boxed{\frac{3}{4}}$.