jmc

Let the side of the octahedron be of length $s$. Let the vertices of the octahedron be $A,B,C,D,E,F$ so that $A$ and $F$ are opposite each other and $AF=s\sqrt{2}$. The height of the square pyramid $ABCDE$ is $\frac{AF}{2}=\frac{s}{\sqrt{2}}$ and so it has volume $\frac{1}{3}{s}^2\cdot \frac{s}{\sqrt{2}}=\frac{s^3}{3\sqrt{2}}$ and the whole octahedron has volume $\frac{s^3\sqrt{2}}{3}$. Let $M$ be the midpoint of $BC$, $N$ be the midpoint of $DE$, $G$ be the centroid of $△ABC$ and $H$ be the centroid of $△ADE$. Then $△AMN\sim △AGH$ and the symmetry ratio is $\frac{2}{3}$ (because the medians of a triangle are trisected by the centroid), so $GH=\frac{2}{3}MN=\frac{2s}{3}$. $GH$ is also a diagonal of the cube, so the cube has side-length $\frac{s\sqrt{2}}{3}$ and volume $\frac{2s^3\sqrt{2}}{27}$. The ratio of the volumes is then $\frac{\left(\frac{2s^3\sqrt{2}}{27}\right)}{\left(\frac{s^3\sqrt{2}}{3}\right)}=\frac{2}{9}$ and so the answer is $\boxed{11}$.