[Solution] A small regular tetrahedron that is tangent to a sphere can be constructed with a plane parallel to the horizontal plane as shown in Figure (1). Let the edge of this tetrahedron be $t$. Then construct the quadrilateral $C_{1} B_{1} B_{2} C_{2}$, where $C_{1}$ and $C_{2}$ are the centers of the two spheres.
According to the symmetry of the sphere and the tetrahedron, the side $C_{1} B_{1} \underline{\Perp} C_{2} B_{2}$, thus
$$
B_{1} B_{2}=C_{1} C_{2}=1+1=2.
$$

This way, $s=t+2$, and our problem reduces to determining $t$.
As shown in Figure (2), in $\triangle A B C_{1}$, the length of $A C_{1}$ is the height $D B$ or $A E$ of the small tetrahedron, $b$, minus the radius of the sphere, and the length of $A B$ is $\frac{2}{3}\left(\frac{\sqrt{3}}{2} t\right)$, because $B$ is the centroid of the tetrahedron's face.
Applying the Pythagorean theorem to $\triangle A B C_{1}$, we get
$$
(b-1)^{2}=1^{2}+\left[\frac{2}{3}\left(\frac{\sqrt{3}}{2} t\right)\right]^{2},
$$

and applying the Pythagorean theorem to $\triangle A D B$, we get
$$
t^{2}=b^{2}+\left[\frac{2}{3}\left(\frac{\sqrt{3}}{2} t\right)\right]^{2}.
$$

Solving the above two equations simultaneously gives the positive value of $t$ as $t=2 \sqrt{6}$.
Thus,
$$
s=2+t=2+2 \sqrt{6}.
$$

Therefore, the answer is $(E)$.
$24 \cdot 12$ Four spheres, each with a radius of 1, are mutually tangent. Three are on the floor, and the fourth is stacked on top of the other three. A regular tetrahedron with edge length $s$ is circumscribed around these spheres. Then $s$ is equal to
(A) $4 \sqrt{2}$.
(B) $4 \sqrt{3}$.
(C) $2 \sqrt{6}$.
(D) $1+2 \sqrt{6}$.
(E) $2+2 \sqrt{6}$.
(31st American High School Mathematics Examination, 1980)