If the three rods start from a single point at right angles, I worked out a box with a base touching all three rods to be:
L=W=(sqrt(3) + 1)/2
H=1/sqrt(3)
Volume=(2+sqrt(3))/(2sqrt(3)) = 1.07735
This is of course larger than a simple cube with a volume of 1. So the
cube is likely the maximum required volume.
If the three rods start from a single point to form three edges of a tetrahedron, I found the box to be:
L=W=(sqrt(3) + 1)/(2sqrt(2))
H=sqrt(2/3)
Volume=(2+sqrt(3))/(2sqrt(6))=.7618
Now, I could be wrong on these...