6. In trapezoid $A B C D$, $A D / / B C, E F$ is the midline, the area ratio of quadrilateral $A E F D$ to quadrilateral $E B C F$ is $\frac{\sqrt{3}+1}{3-\sqrt{3}}$, and the area of $\triangle A B D$ is $\sqrt{3}$. Then the area of trapezoid $A B C D$ is
6. 2 .

Let $A D=a, B C=b$, and the height be $2 h$. The area of trapezoid $A B C D$ is $S$. Then
$$
\begin{array}{l}
E F=\frac{a+b}{2}, \frac{a}{b}=\frac{S_{\triangle A B D}}{S_{\triangle B C D}}=\frac{\sqrt{3}}{S-\sqrt{3}} . \\
\text { Also, } \frac{\sqrt{3}+1}{3-\sqrt{3}}=\frac{S_{\text {quadrilateral } A E F D}}{S_{\text {quadrilateral } E B C F}}=\frac{S+2 \sqrt{3}}{3 S-2 \sqrt{3}} .
\end{array}
$$

Applying the ratio theorem, we get $S=2$.