6. Given planar vectors $\boldsymbol{a} \boldsymbol{b} 、 \boldsymbol{c}$ satisfy
$$
|a|=|b|=a \cdot b=c \cdot(a+2 b-2 c)=2 \text {. }
$$

If the maximum and minimum values of $|\boldsymbol{a}-\boldsymbol{c}|$ are $M$ and $m$, respectively, then $\frac{M}{m}=$ $\qquad$
6. $\frac{5+\sqrt{21}}{2}$.

Given $|\boldsymbol{a}|=|\boldsymbol{b}|=\boldsymbol{a} \cdot \boldsymbol{b}=2$, we know $\langle\boldsymbol{a}, \boldsymbol{b}\rangle=60^{\circ}$.
Establish a rectangular coordinate system with the origin at the starting point of $\boldsymbol{a}$ and the direction of $\boldsymbol{a}$ as the $x$-axis.
Thus, $\boldsymbol{a}=(2,0), \boldsymbol{b}=(1, \sqrt{3})$.
Let $\boldsymbol{c}=(x, y)$. Then
$$
\begin{array}{l}
\boldsymbol{c} \cdot(\boldsymbol{a}+2 \boldsymbol{b}-2 \boldsymbol{c})=4 x+2 \sqrt{3} y-2 x^{2}-2 y^{2}=2 \\
\Rightarrow(x-1)^{2}+\left(y-\frac{\sqrt{3}}{2}\right)^{2}=\frac{3}{4} .
\end{array}
$$

Therefore, $|\boldsymbol{a}-\boldsymbol{c}|=\sqrt{(x-2)^{2}+y^{2}}$
$$
\in\left[\frac{\sqrt{7}-\sqrt{3}}{2}, \frac{\sqrt{7}+\sqrt{3}}{2}\right] \text {. }
$$

Thus, $\frac{M}{m}=\frac{5+\sqrt{21}}{2}$.