A regular octahedron is formed by joining the centers of adjoining faces of a cube. The ratio of the volume of the octahedron to the volume of the cube is

$ \textbf{(A)}\ \frac{\sqrt{3}}{12} \qquad\textbf{(B)}\ \frac{\sqrt{6}}{16} \qquad\textbf{(C)}\ \frac{1}{6} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{8} \qquad\textbf{(E)}\ \frac{1}{4} $
1. **Identify the side length of the cube and the octahedron:**
   Let the side length of the cube be \( a \). The octahedron is formed by joining the centers of the adjoining faces of the cube. The distance between the centers of two adjacent faces of the cube is \( \frac{a}{\sqrt{2}} \). This distance is the edge length of the octahedron.

2. **Calculate the volume of the cube:**
   The volume \( V_{\text{cube}} \) of the cube with side length \( a \) is given by:
   \[
   V_{\text{cube}} = a^3
   \]

3. **Calculate the volume of the octahedron:**
   The volume \( V_{\text{octahedron}} \) of a regular octahedron with edge length \( b \) is given by:
   \[
   V_{\text{octahedron}} = \frac{\sqrt{2}}{3} b^3
   \]
   Here, \( b = \frac{a}{\sqrt{2}} \).

4. **Substitute \( b \) into the volume formula for the octahedron:**
   \[
   V_{\text{octahedron}} = \frac{\sqrt{2}}{3} \left( \frac{a}{\sqrt{2}} \right)^3
   \]
   Simplify the expression inside the parentheses:
   \[
   \left( \frac{a}{\sqrt{2}} \right)^3 = \frac{a^3}{(\sqrt{2})^3} = \frac{a^3}{2\sqrt{2}}
   \]
   Therefore:
   \[
   V_{\text{octahedron}} = \frac{\sqrt{2}}{3} \cdot \frac{a^3}{2\sqrt{2}} = \frac{\sqrt{2} \cdot a^3}{3 \cdot 2\sqrt{2}} = \frac{a^3}{6}
   \]

5. **Calculate the ratio of the volumes:**
   The ratio of the volume of the octahedron to the volume of the cube is:
   \[
   \text{Volume ratio} = \frac{V_{\text{octahedron}}}{V_{\text{cube}}} = \frac{\frac{a^3}{6}}{a^3} = \frac{1}{6}
   \]

The final answer is \(\boxed{\frac{1}{6}}\).