6. In a regular tetrahedron with a base edge length of 4 and a height of 2, first place a sphere \( B_{1} \) that is tangent to the base and all three side faces. Then, for \( n=2,3, \cdots \), place spheres \( B_{n} \) in the regular tetrahedron such that they are externally tangent to \( B_{n-1} \) and tangent to all three side faces of the tetrahedron. The sum of the volumes of all these spheres is
6. $\frac{16 \pi}{39}$.

Let the radius of sphere $B_{n}$ be $r_{n}$.
Given that the base edge length of the regular tetrahedron is 4, and the height is 2, it is easy to calculate that the radius of the inscribed sphere $B_{1}$ is $r_{1}=\frac{2}{3}$.

Draw a tangent plane through the tangent points of spheres $B_{1}$ and $B_{2}$, then a smaller tetrahedron is separated, which is similar to the larger tetrahedron, and $B_{2}$ is the inscribed sphere of the smaller tetrahedron.

Since the height of the smaller tetrahedron is $2-2 r_{1}=\frac{2}{3}$, the similarity ratio is $\frac{2}{3}: 2=1: 3$.
Thus, $r_{2}: r_{1}=1: 3$.
Similarly,
$$
r_{2}: r_{1}=r_{3}: r_{2}=\cdots=r_{n}: r_{n-1}=1: 3 .
$$

Therefore, the sum of the volumes is
$$
\frac{4 \pi}{3}\left(r_{1}^{3}+r_{2}^{3}+\cdots\right)=\frac{4 \pi}{3} \times \frac{\left(\frac{2}{3}\right)^{3}}{1-\left(\frac{1}{3}\right)^{3}}=\frac{16 \pi}{39} .
$$