7. As shown in Figure 2, the area of the square $D E F G$ inscribed in the equilateral triangle $\triangle A B C$ with area 1 is $a \sqrt{b}-c$, where $a, b, c$ are integers, and $b$ cannot be divided by the square of any prime number. Then $\frac{a-c}{b}$
7. $-\frac{20}{3}$.

Let the side length of square DEFG be $x$, and the side length of equilateral $\triangle ABC$ be $m$. Then $m^{2}=\frac{4}{\sqrt{3}}$.
From $\triangle ADG \backsim \triangle ABC$, we get
$$
\frac{x}{m}=\frac{\frac{\sqrt{3}}{2} m-x}{\frac{\sqrt{3}}{2} m} .
$$

Solving this, we get $x=(2 \sqrt{3}-3) m$.
Thus, $x^{2}=(2 \sqrt{3}-3)^{2} m^{2}=28 \sqrt{3}-48$.
By the problem, $a=28, b=3, c=48$. Therefore, $\frac{a-c}{b}=-\frac{20}{3}$.