6. Given 4 spheres with a radius of 1, each pair of which are externally tangent, then the edge length of the circumscribed regular tetrahedron of the geometric body formed by these 4 spheres is
As shown in the figure, $O_{1} O_{2} O_{3} O_{4}$ is a regular tetrahedron with a side length of 2, and its height is $h_{1}=\frac{\sqrt{6}}{3} \times 2=\frac{2 \sqrt{6}}{3}$.
Obviously, the center of the regular tetrahedron $O_{1} O_{2} O_{3} O_{4}$ coincides with the center of the circumscribed regular tetrahedron $A B C D$, which is denoted as $O$.
Then the distance from $O$ to the base $A B C$ is $h_{2}=\frac{1}{4} h_{1}+1$
$=\frac{\sqrt{6}}{6}+1 \Rightarrow$ the height of the regular tetrahedron $A B C D$ is
$$
h=4 h_{2}=\frac{2 \sqrt{6}}{3}+4 .
$$

Therefore, the edge length of the regular tetrahedron $A B C D$ is $a=\frac{\sqrt{6}}{2} h=2 \sqrt{6}+2$.