6. It is known that a regular tetrahedron can accommodate 10 small spheres with a radius of 1. Then the minimum edge length of the regular tetrahedron is $\qquad$
6. $4+2 \sqrt{6}$.

When the edge length of a regular tetrahedron is minimized, let the edge length be $a$. At this time, the first, second, and third layers have $1$, $3$, and $6$ balls, respectively, and adjacent balls are tangent to each other.
Note that, the centroid divides the height of the tetrahedron in the ratio $1: 3$.
Thus, the height of the regular tetrahedron is
$$
\begin{array}{l}
h=\sqrt{\frac{2}{3}} a=3+\sqrt{\frac{2}{3}} \times 2+\sqrt{\frac{2}{3}} \times 2+1 \\
\Rightarrow a=4+2 \sqrt{6} .
\end{array}
$$