Solution:

Without loss of generality, we may assume that the original hexagon has side length $3$. Let $s$ then be the side length of the inscribed hexagon. Notice that there are six triangles with sides $1$, $2$, and $s$, and the angle between the sides of lengths $1$ and $2$ is $120^{\circ}$, as shown below:

By the cosine law, we have
$$s^2 = 1^2 + 2^2 - 2 \cdot 1 \cdot 2 \cdot \left(-\frac{1}{2}\right) = 7$$
and so $s = \sqrt{7}$. Since the ratio of the areas of two similar shapes is equal to the square of the ratio of their sides, the desired ratio is thus
$$\left(\frac{\sqrt{7}}{3}\right)^2 = \frac{7}{9}.$$