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
$$\begin{aligned} & A C = 2 R_{\Gamma} \cdot \sin 60^{\circ} = 3 \sqrt{3} \\ & A E = 2 R_{\gamma} \cdot \sin 60^{\circ} = 2 \sqrt{3} \end{aligned}$$
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 $B D = B E + E D = \frac{9 \sqrt{21}}{7}$.