We are clearly looking for the regular tetrahedron inscribed in a sphere of radius 1 (i.e. with all its vertices lying on the sphere). The neat trick with regular tetrahedra is to inscribe them in a cube.
Wikipedia has a picture of the two regular tetrahedra you can find in a cube: http://en.wikipedia.org/wiki/File:CubeAndStel.gif
The cube inscribed in a unit sphere has side length $\frac{2}{\sqrt 3}$, so the regular tetrahedron has side length $x = \sqrt{2} \frac{2}{\sqrt 3} = \sqrt{\frac{8}{3}}$.
Here is another example of this idea: Height of a tetrahedron
(In this answer, I do as the questioner did, calling the smallest containing sphere the circumsphere and its center the circumcenter, even if this sphere does not touch all vertices. Similarly, I call the largest enclosed sphere the insphere and its center the incenter, even if this sphere does not touch all faces.)
Assuming that the circumcenters and incenters of both the original polyhedron and the dual all coincide, these ratios should be the same.
A convex polyhedron which contains the origin in its interior can be described either as the convex hull of a set $\{a_i\}\subseteq {\Bbb R}^3$ of vertices or as the intersection of a set of bounding half-spaces $\{S_j\}$, where each bounding half-space is of the form
$$S_j=\{x\mid x\cdot b_j\le 1\}, \qquad b_j\in {\Bbb R}^3. $$
Letting the circumsphere be centered at the origin, the circumradius $R$ will equal the maximum value of any $|a_i|$. Also, since the distance between the origin and the plane $\{x\mid x\cdot b_j=1\}$ is $|b_j|^{-1}$, if the insphere is centered at the origin, the inradius will equal the minimum value of any $|b_j|^{-1}$.
The polar dual of a given polyhedron can be constructed by interchanging the $a_i$s and $b_j$s. So, if the original polyhedron had circumradius $R$ and inradius $r$, and if the circumcenter and incenter of the polar dual are still at the origin, then the polar dual must have circumradius $1/r$ and inradius $1/R$, and the ratio $R/r$ of the circumradius to the inradius remains the same.
For an example, take a cuboctahedron with vertices
$$
(\pm 1, \pm 1, 0), \ (\pm 1, 0, \pm 1), \ (0, \pm 1, \pm 1).
$$
In this case, the circumsphere and insphere are both centered at the origin. The radius $\sqrt{2}$ circumsphere touches all vertices; the insphere has radius $1$ and touches only the square faces; and the ratio $R/r$ is $\sqrt{2}$.
The dual of this polyhedron is the rhombic dodecahedron, with vertices
$$
(\pm 1,0,0), \ (0,\pm 1,0), \ (0,0,\pm 1),\ (\pm\frac12, \pm\frac12, \pm\frac12).
$$
The circumsphere and the insphere are both still centered at the origin. The circumsphere has radius $1$ and touches only those vertices where four faces meet (the first six above); the insphere has radius $1/\sqrt{2}$ and touches all faces; and the ratio $R/r$ is still $\sqrt{2}$.
Best Answer
Assuming the $\text{centroid}\equiv\text{incenter}\equiv\text{circumcenter}$ of the regular tetrahedron $ABCD$ lies at the origin, i.e. $A+B+C+D=0$, the ratio $\frac{R}{r}$ equals $\frac{OA}{OA'}$ with $A'$ being the centroid of $BCD$. Hence $$ \frac{R}{r} = \frac{3\|A\|}{\|B+C+D\|} =\frac{3\|A\|}{\|A\|} = 3 $$ with a straightforward generalization to the regular $n$-simplex.
This can be shown also by embedding $ABCD$ in $\mathbb{R}^4$ via $A\mapsto(1,0,0,0),B\mapsto(0,1,0,0),$ $C\mapsto(0,0,1,0),$ $D\mapsto(0,0,0,1)$. Metric computations in this framework are utterly simple.