6. As shown in Figure $1, M$ and $N$ are internal points on the diagonals $A C$ and $C E$ of the regular hexagon $A B C D E F$, respectively, and $\frac{A M}{A C}=\frac{C N}{C E}=\lambda$. If points $B$, $M$, and $N$ are collinear, then $\lambda=$ ( ).
(A) $\frac{\sqrt{3}}{3}$
(B) $\frac{1}{3}$
(C) $\frac{\sqrt{2}}{2}$
(D) $\frac{1}{2}$
6. A.

Set $B$ as the origin and the line $B C$ as the $x$-axis to establish a rectangular coordinate system.
Assume $B C=1$. Then
$$
A\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right), C(1,0), E(1, \sqrt{3}) \text {. }
$$

Given $\frac{A M}{A C}=\frac{C N}{C E}=\lambda$, using the section formula, we get
$$
M\left(\frac{3}{2} \lambda-\frac{1}{2}, \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2} \lambda\right), N(1, \sqrt{3} \lambda) .
$$

Since $B$, $M$, and $N$ are collinear, we have
$$
\frac{\sqrt{3} \lambda}{1}=\frac{\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2} \lambda}{\frac{3}{2} \lambda-\frac{1}{2}} \Rightarrow \lambda= \pm \frac{\sqrt{3}}{3} \text {. }
$$

Noting that $\lambda>0$, we have $\lambda=\frac{\sqrt{3}}{3}$.