Six circles form a ring with with each circle externally tangent to two circles adjacent to it. All circles are internally tangent to a circle $C$ with radius $30$. Let $K$ be the area of the region inside circle $C$ and outside of the six circles in the ring. Find $\lfloor K \rfloor$.
1. **Understanding the Geometry**:
   - We have six smaller circles, each with radius \( r \), forming a ring.
   - Each smaller circle is externally tangent to its two adjacent smaller circles.
   - All smaller circles are internally tangent to a larger circle \( C \) with radius \( 30 \).

2. **Finding the Radius of the Smaller Circles**:
   - Consider the center of the larger circle \( C \) and the center of one of the smaller circles.
   - The distance from the center of the larger circle to the center of a smaller circle is \( 30 - r \).
   - The centers of two adjacent smaller circles form an equilateral triangle with side length \( 2r \) (since they are tangent to each other).

3. **Using Geometry to Solve for \( r \)**:
   - The distance from the center of the larger circle to the center of a smaller circle is also the radius of the larger circle minus the radius of the smaller circle, i.e., \( 30 - r \).
   - The centers of the smaller circles form a regular hexagon inscribed in a circle of radius \( 30 - r \).
   - The side length of this hexagon is \( 2r \).

4. **Relating the Side Length to the Radius**:
   - The radius of the circumscribed circle of a regular hexagon is equal to the side length of the hexagon.
   - Therefore, \( 30 - r = 2r \).

5. **Solving for \( r \)**:
   \[
   30 - r = 2r
   \]
   \[
   30 = 3r
   \]
   \[
   r = 10
   \]

6. **Calculating the Area**:
   - The area of the larger circle \( C \) is:
     \[
     \pi (30)^2 = 900\pi
     \]
   - The area of one smaller circle is:
     \[
     \pi (10)^2 = 100\pi
     \]
   - The total area of the six smaller circles is:
     \[
     6 \cdot 100\pi = 600\pi
     \]

7. **Finding the Area of the Region Inside Circle \( C \) and Outside the Six Smaller Circles**:
   \[
   K = 900\pi - 600\pi = 300\pi
   \]

8. **Floor Function**:
   - We need to find \( \lfloor K \rfloor \):
     \[
     \lfloor 300\pi \rfloor
     \]
   - Using the approximation \( \pi \approx 3.141592653589793 \):
     \[
     300\pi \approx 300 \times 3.141592653589793 = 942.4777960769379
     \]
   - Therefore, \( \lfloor 300\pi \rfloor = 942 \).

The final answer is \(\boxed{942}\).