Solution: $5 \sqrt{2} / 3$
We can construct a cube such that the vertices of the cuboctahedron are the midpoints of the edges of the cube.

Let $s$ be the side length of this cube. Now, the cuboctahedron is obtained from the cube by cutting a tetrahedron from each corner. Each such tetrahedron has a base in the form of an isosceles right triangle of area $(s / 2)^{2} / 2$ and height $s / 2$ for a volume of $(s / 2)^{3} / 6$. The total volume of the cuboctahedron is therefore
$$s^{3}-8 \cdot(s / 2)^{3} / 6=5 s^{3} / 6$$
Now, the side of the cuboctahedron is the hypotenuse of an isosceles right triangle of leg $s / 2$; thus $1=(s / 2) \sqrt{2}$, giving $s=\sqrt{2}$, so the volume of the cuboctahedron is $5 \sqrt{2} / 3$.