II. (25 points) As shown in Figure 6, with the vertices of an equilateral triangle $\triangle ABC$ with side length 2 as centers and the side length as the radius, draw three equal circles, obtaining intersection points $D, E, F$. Connect $CF$, intersecting $\odot C$ at point $G$. With $E$ as the center and $EG$ as the radius, draw an arc intersecting sides $AB$ and $BC$ at points $M$ and $N$, respectively. Connect $MN$. Find the inradius $r$ of $\triangle BMN$.
II. As shown in Figure 8, draw $EP \perp AB$, intersecting the extension of $BA$ at point $P$, and connect $EA$, $EC$, $EG$, and $EM$.
It is easy to see that $\triangle EAC$ is an equilateral triangle. Therefore, $EC \parallel AB$.
Since $CG \perp AB \Rightarrow EC \perp CG$
$\Rightarrow EM = EG = 2\sqrt{2}$.
Given $\angle EAP = 60^\circ \Rightarrow EP = \sqrt{3}, AP = 1$.
Thus, $PM = \sqrt{EM^2 - EP^2} = \sqrt{5}$
$\Rightarrow AM = PM - AP = \sqrt{5} - 1$
$\Rightarrow BM = AB - AM = 2 - (\sqrt{5} - 1) = 3 - \sqrt{5}$.
It is easy to see that points $M$ and $N$ are symmetric with respect to line $BE$.
Therefore, $BN = BM = 3 - \sqrt{5}$, and $MN \parallel AC$.
Thus, $\triangle BMN$ is an equilateral triangle.
Hence, $r = \frac{BN}{2} \tan 30^\circ = \frac{3\sqrt{3} - \sqrt{15}}{6}$.