Answer:
$\frac{9 \sqrt{21}}{7}$
Solution: The key observation is that $\triangle A C D$ is equilateral. This is proven in two steps.
- From tangency at $C$, we have
$$
\angle D C A=\angle D C E=\angle E B C=\angle D B C=\angle D A C,
$$
implying that $C A=C D$.
- Consider the common tangent of $\gamma$ and $\Gamma$ at $A$. By homothety at $E$, this line is parallel to the tangent of $\odot(E B C)$ at $C$, which is line $C D$. This implies that $A C=A D$.

Once we have this, compute
$$
\begin{array}{l}
A C=2 R_{\Gamma} \cdot \sin 60^{\circ}=3 \sqrt{3} \\
A E=2 R_{\gamma} \cdot \sin 60^{\circ}=2 \sqrt{3}
\end{array}
$$

There are now many ways to finish. One way is to use Stewart's theorem on $\triangle A D C$ to get $E D=\sqrt{21}$, then use Power of Point to get $E B=\frac{A E \cdot E C}{E D}=\frac{2 \sqrt{21}}{7}$. The final answer is $B D=B E+E D=\frac{9 \sqrt{21}}{7}$.
7. Quadrilateral $A B C D$ is inscribed in circle $\Gamma$. Segments $A C$ and $B D$ intersect at $E$. Circle $\gamma$ passes through $E$ and is tangent to $\Gamma$ at $A$. Suppose that the circumcircle of triangle $B C E$ is tangent to $\gamma$ at $E$ and is tangent to line $C D$ at $C$. Suppose that $\Gamma$ has radius 3 and $\gamma$ has radius 2. Compute $B D$.