6. For the arithmetic sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, the sums of the first $n$ terms are $S_{n}$ and $T_{n}$, respectively. If for any positive integer $n$, we have
$$
\begin{array}{r}
\frac{S_{n}}{T_{n}}=\frac{5 n-3}{2 n+1}, \\
\text { then } \frac{a_{20}}{b_{7}}=
\end{array}
$$
6. $\frac{64}{9}$.

Since $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are both arithmetic sequences, we set
$$
S_{n}=k n(5 n-3), T_{n}=k n(2 n+1) \text {. }
$$

When $n \geqslant 2$,
$$
\begin{array}{l}
a_{n}=S_{n}-S_{n-1}=k(10 n-8), \\
b_{n}=T_{n}-T_{n-1}=k(4 n-1) .
\end{array}
$$

Thus, $\frac{a_{20}}{b_{7}}=\frac{k(200-8)}{k(28-1)}=\frac{64}{9}$.