Two cubes $A$ and $B$ have different side lengths, such that the volume of cube $A$ is numerically equal to the surface area of cube $B$. If the surface area of cube $A$ is numerically equal to six times the side length of cube $B$, what is the ratio of the surface area of cube $A$ to the volume of cube $B$?
1. Let \( a \) denote the side length of cube \( A \), and \( b \) denote the side length of cube \( B \).

2. We are given that the volume of cube \( A \) is numerically equal to the surface area of cube \( B \). The volume of cube \( A \) is \( a^3 \), and the surface area of cube \( B \) is \( 6b^2 \). Therefore, we have:
   \[
   a^3 = 6b^2
   \]

3. We are also given that the surface area of cube \( A \) is numerically equal to six times the side length of cube \( B \). The surface area of cube \( A \) is \( 6a^2 \), and six times the side length of cube \( B \) is \( 6b \). Therefore, we have:
   \[
   6a^2 = 6b
   \]
   Simplifying this equation, we get:
   \[
   a^2 = b
   \]

4. Substitute \( b = a^2 \) into the first equation \( a^3 = 6b^2 \):
   \[
   a^3 = 6(a^2)^2
   \]
   Simplifying, we get:
   \[
   a^3 = 6a^4
   \]
   Dividing both sides by \( a^3 \) (assuming \( a \neq 0 \)), we get:
   \[
   1 = 6a
   \]
   Therefore:
   \[
   a = \frac{1}{6}
   \]

5. Using \( a = \frac{1}{6} \), we find \( b \):
   \[
   b = a^2 = \left(\frac{1}{6}\right)^2 = \frac{1}{36}
   \]

6. We need to find the ratio of the surface area of cube \( A \) to the volume of cube \( B \). The surface area of cube \( A \) is:
   \[
   6a^2 = 6 \left(\frac{1}{6}\right)^2 = 6 \cdot \frac{1}{36} = \frac{1}{6}
   \]
   The volume of cube \( B \) is:
   \[
   b^3 = \left(\frac{1}{36}\right)^3 = \frac{1}{36^3} = \frac{1}{46656}
   \]

7. The ratio of the surface area of cube \( A \) to the volume of cube \( B \) is:
   \[
   \frac{\frac{1}{6}}{\frac{1}{46656}} = \frac{1}{6} \cdot 46656 = 7776
   \]

The final answer is \(\boxed{7776}\).