jmc

Let $r$ be the radius of each of the six congruent circles, and let $A$ and $B$ be the centers of two adjacent circles. Join the centers of adjacent circles to form a regular hexagon with side $2r$. Let $O$ be the center of $\mathcal{C}$. Draw the radii of $\mathcal{C}$ that contain $A$ and $B$. Triangle $ABO$ is equilateral, so $OA=OB=2r$. Because each of the two radii contains the point where the smaller circle is tangent to $\mathcal{C}$, the radius of $\mathcal{C}$ is $3r$, and $K=\pi \left((3r{)}^2-6r^2\right)=3\pi r^2$. The radius of $\mathcal{C}$ is 30, so $r=10$, $K=300\pi$, and $\lfloor K\rfloor =\boxed{942}$.