Consider a regular octahedron $A$-$BCDE$-$F$. Denote by $\pi_1$, $\pi_2$ and $\pi_3$ the planes passing through the diagonals $BD$ and $CE$, the diagonals $CE$ and $AF$, and the diagonals $BD$ and $EF$, respectively.
By symmetry, $\pi_1$, $\pi_2$ and $\pi_3$ divide the octahedron into 8 equal parts. The octahedron is divided into 8 triangular pyramids, and the cube is divided into 8 small cubes. The ratio of the volume of the small cube and the triangular pyramid is equal to that of the original cube and the octahedron. We will calculate the volume ratio of the small cube and the triangular pyramid.

Call $O$ the intersection of the 3 diagonals. Consider the triangular pyramid $OABC$ and the cube in it. $\angle AOB$, $\angle BOC$ and $\angle COA$ are right angles. Let $M$ be the midpoint of $BC$ and let $G$ be the centroid of the triangle $ABC$. Then $O$ and $G$ are vertices of the small cube.

Since the centroid of a triangle divides a median in the ratio $2 : 1$, $A$ is three times as high as $G$ is from the plane $OBC$. Let $OA = x$. The length of the edges of the small cube is $\frac{1}{3}x$ and so its volume is $\frac{1}{27}x^3$. The volume of the triangular pyramid is $\frac{1}{3} \cdot x \cdot \frac{1}{2}x^2 = \frac{1}{6}x^3$. Thus, the ratio of the volume of the small cube and that of the triangular pyramid is $\frac{1}{27}x^3 \div \frac{1}{6}x^3 = \frac{2}{9}$.