6. In $\triangle A B C$, $A B=1, A C=2, \cos B=2 \sin C$, then the length of $B C$ is
$$
\begin{array}{l}
\frac{c}{\sin C}=\frac{b}{\sin B} \Rightarrow \frac{2}{\cos B}=\frac{2}{\sin B} \Rightarrow B=\frac{\pi}{4}, \text { then } b^{2}=a^{2}+c^{2}-2 a c \cos B \\
\Rightarrow 4=a^{2}+1-\sqrt{2} a \Rightarrow a^{2}-\sqrt{2} a-3=0 \Rightarrow a=\frac{\sqrt{2}+\sqrt{14}}{2} .
\end{array}
$$