6. In the geometric sequence $\left\{a_{n}\right\}$, $a_{1}=\frac{1}{8}$, the geometric mean of the first $n$ terms is 8. If the geometric mean of the remaining terms after removing one term from the first $n$ terms is $4 \sqrt{2}$, then the removed term is the $\qquad$th term.
6.13.

Let the common ratio be $q$, the product of the first $n$ terms be $M$, and the $k$-th term be removed, then $M=8^{n}$, and $\frac{M}{a_{k}}=(4 \sqrt{2})^{n-1}$.
Thus, $a_{k}=(\sqrt{2})^{n+5}$, which means $\frac{1}{8} q^{k-1}=(\sqrt{2})^{n+5}$.
Therefore, $q^{k-1}=(\sqrt{2})^{n+11}$.
Also, $M=8^{n}$, i.e., $\left(\frac{1}{8}\right)^{n} q^{\frac{n(n-1)}{2}}=8^{n}$, which gives $q^{n-1}=(\sqrt{2})^{24}$.
Comparing (1) and (2), we have $k-1=\frac{(n-1)(n+11)}{24}$.
Given $k \leqslant n$, we solve to get $n \leqslant 13$, and
$k-1=\frac{(n-1)^{2}+12(n-1)}{24}$.
Thus, 12 divides $n-1$.
Therefore, $n=1$ (contradicts the given) or $n=13$. Hence, $k=13$.