6. Given a regular quadrilateral prism $A B C D-A_{1} B_{1} C_{1} D_{1}$ with a base edge length of 2 and a side edge length of $4, E$ is the midpoint of edge $C D$, and $F$ is the midpoint of edge $A A_{1}$. Then, the distance from point $D$ to the plane $E F B_{1}$ is $\qquad$.
Answer
As shown in the figure, let the cross-section be $E G B_{1} F H, A_{1} F=A_{1} B_{1}=2$
$$
\Rightarrow C G=C E=1 \Rightarrow C_{1} G=3 \text {. }
$$

Then $\triangle F A H \sim \triangle G C_{1} B_{1} \Rightarrow A H=\frac{4}{3}, D H=\frac{2}{3}$.
Thus, $H E=\frac{\sqrt{13}}{3}, H F=\frac{2 \sqrt{13}}{3}, E F=3$,
so $\cos \angle E H F=-\frac{4}{13} \Rightarrow \sin \angle E H F=\frac{3 \sqrt{17}}{13}$
$\Rightarrow S_{\triangle E H F}=\frac{1}{2} \cdot \frac{\sqrt{13}}{3} \cdot \frac{2 \sqrt{13}}{3} \cdot \frac{3 \sqrt{17}}{13}=\frac{\sqrt{17}}{3}$.
Also, $V_{D-E H F}=V_{F-H D E}=\frac{1}{3} \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot 1 \cdot 2=\frac{2}{9}$,
Let the distance from point $D$ to the plane $E F B_{1}$ be $h$,
Therefore, $V_{D-E H F}=\frac{1}{3} \cdot \frac{\sqrt{17}}{3} \cdot h=\frac{2}{9} \Rightarrow h=\frac{2}{\sqrt{17}}=\frac{2 \sqrt{17}}{17}$.