6. Let the edge lengths of three regular tetrahedra be integers, and the sum of their volumes is $\frac{51 \sqrt{2}}{4}$. Then the sum of the surface areas of these regular tetrahedra is $\qquad$ .
6. $35 \sqrt{3}$.

Let the edge lengths of three regular tetrahedra be $a, b, c$ respectively.
Assume $0 < a \leqslant b \leqslant c$.
$$
\begin{array}{l}
\text { By } \frac{\sqrt{2}}{12}\left(a^{3}+b^{3}+c^{3}\right)=\frac{51 \sqrt{2}}{4} \\
\Rightarrow a^{3}+b^{3}+c^{3}=153,
\end{array}
$$

Here, $c^{3} \leqslant 125$, i.e., $c \leqslant 5$.
Since $3 c^{3} \geqslant a^{3}+b^{3}+c^{3}=153$, thus, $c^{3} \geqslant 51 \Rightarrow c \geqslant 4$.
Therefore, $c=4$ or $c=5$.
If $c=4$, there are no suitable $a, b$;
If $c=5$, it is easy to find $a=1, b=3$.
Thus, the sum of the surface areas is
$$
\begin{array}{l}
4\left(\frac{1}{2} a^{2} \sin 60^{\circ}+\frac{1}{2} b^{2} \sin 60^{\circ}+\frac{1}{2} c^{2} \sin 60^{\circ}\right) \\
=\sqrt{3}\left(a^{2}+b^{2}+c^{2}\right)=\sqrt{3}\left(1^{2}+3^{2}+5^{2}\right) \\
=35 \sqrt{3} .
\end{array}
$$