This is a Truncated Tetrahedron (8 faces: 4 regular hexagons and 4 equilateral triangles) included into a 6x6x6 unit cube.
This huge (more than 6x6x6cm) but still affordable model demonstrates an interesting result.
I shows that, since the coodinates of the vertices are of the form (3,1,1) with permutations or minus signs, the distance from the center of the polyhedron to a vertex is sqrt(3x3 + 1x1 + 1x1) i.e. sqrt(11), while the size of a side is 2sqrt(2), the diagonal of two squares.
Thus a truncated tetrahedron whose vertices are located on the unit sphere would have a side of 2sqrt(2/11). Nice result, just looking at a model...
But there is more: with a similar argument, the separation between two vertices, that is the angle theta between two segments going from the center of the polyhedron to two vertices that are neighbours can be found by this relationship:
sin(thetha/2) = sqrt(2/11), which gives theta=50.4788°
6.202 w x
6.2 d x
2.442 w x
2.441 d x