imc

We can break the octahedron into two square pyramids by cutting it along a plane perpendicular to one of its internal diagonals. The cube has edges of length 1 so all edges of the regular octahedron have length $\frac{\sqrt{2}}{2}$. Then the square base of the pyramid has area ${\left(\frac{1}{2}\sqrt{2}\right)}^2=\frac{1}{2}$. We also know that the height of the pyramid is half the height of the cube, so it is $\frac{1}{2}$. The volume of a pyramid with base area $B$ and height $h$ is $A=\frac{1}{3}Bh$ so each of the pyramids has volume $\frac{1}{3}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{12}$. The whole octahedron is twice this volume, so $\frac{1}{12}\cdot 2=\boxed{\frac{1}{6}}$.