6. $6 \mathrm{~cm}$.

Solution: Let the ratio of the area of side $A C C_{1} A_{1}$ to the area of side $A B B_{1} A_{1}$ be $5: 8, D E F$ be a direct cross-section of the oblique triangular prism, then the plane angle of the dihedral angle formed by side $A C C_{1} A_{1}$ and side $A B B_{1} A_{1}$ is $\angle E D F=60^{\circ}$. We can set $D F=5 x, D E=8 x, A A_{1}=l$, in $\triangle D E F$,
$$
\begin{array}{l}
E F^{2}=D F^{2}+D E^{2}-2 D E \cdot D F \cdot \cos 60^{\circ}=25 x^{2} \\
+64 x^{2}-2 \cdot 5 \cdot 8 \cdot \frac{1}{2} x^{2}=49 x^{2} .
\end{array}
$$

So $E F=7 x$, by the given
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ S _ { \text { clear } } = ( 5 x + 8 x + 7 x ) l } \\
{ V = \frac { 1 } { 2 } \cdot 5 x \cdot 8 x \operatorname { s i n } 6 0 ^ { \circ } l }
\end{array} \Rightarrow \left\{\begin{array}{l}
20 x l=60 \\
10 \sqrt{3} x l=15 \sqrt{3}
\end{array} \Rightarrow\right.\right. \\
\left\{\begin{array}{l}
x=\frac{1}{2} \text {, i.e., } l=6 \mathrm{~cm} . \\
l=6
\end{array} \Rightarrow\right.
\end{array}
$$
6. The ratio of the areas of two lateral faces of an oblique triangular prism is $5: 8$, the dihedral angle between them is $60^{\circ}$, the lateral surface area of the prism is 60 $\mathrm{cm}^{2}$, and its volume is $15 \sqrt{3} \mathrm{~cm}$, then the length of its lateral edge is $\qquad$