如图，连接$CE$，$OC$，

由三角形任意两边之差小于第三边得，当$O$、$C$、$E$共线时$OE$最小，
设$\widehat {AC}$的弧度为$x^{\circ}$，
$\therefore \widehat {BC}$的弧度为：$\left(180-x\right)^{\circ}$，
$\because \angle CAD=\angle CAB$，
$\therefore \widehat {CD}$的弧度为：$\left(180-x\right)^{\circ}$，
由折叠得，$\widehat {CDA}$的弧度为$x^{\circ}$，
$\therefore \widehat {AD}$的弧度为：$x^{\circ}-\left(180-x\right)^{\circ}=\left(2x-180\right)^{\circ}$，
$\because$点$E$为弧$AD$中点，
$\therefore \widehat {DE}$的弧度为：$\frac{1}{2}\times \left(2x-180\right)^{\circ}=\left(x-90\right)^{\circ}$，
$\therefore \widehat {CE}$的弧度为：$\left(180-x\right)^{\circ}+\left(x-90\right)^{\circ}=90^{\circ}$，
即$\widehat {CE}$所对圆心角为$90^{\circ}$，
$\because AB=6$，
$\therefore \odot O$半径为$3$，
$\therefore CE=\sqrt{{3}^{2}+{3}^{2}}=3\sqrt{2}$，
$\therefore OE=CE-OC=3\sqrt{2}-3$.
故答案为：$3\sqrt{2}-3$.

如图，连接$CE$，$OC$，

由三角形任意两边之差小于第三边得，当$O$、$C$、$E$共线时$OE$最小，
设$\widehat {AC}$的弧度为$x^{\circ}$，
$\therefore \widehat {BC}$的弧度为：$\left(180-x\right)^{\circ}$，
$\because \angle CAD=\angle CAB$，
$\therefore \widehat {CD}$的弧度为：$\left(180-x\right)^{\circ}$，
由折叠得，$\widehat {CDA}$的弧度为$x^{\circ}$，
$\therefore \widehat {AD}$的弧度为：$x^{\circ}-\left(180-x\right)^{\circ}=\left(2x-180\right)^{\circ}$，
$\because$点$E$为弧$AD$中点，
$\therefore \widehat {DE}$的弧度为：$\frac{1}{2}\times \left(2x-180\right)^{\circ}=\left(x-90\right)^{\circ}$，
$\therefore \widehat {CE}$的弧度为：$\left(180-x\right)^{\circ}+\left(x-90\right)^{\circ}=90^{\circ}$，
即$\widehat {CE}$所对圆心角为$90^{\circ}$，
$\because AB=6$，
$\therefore \odot O$半径为$3$，
$\therefore CE=\sqrt{{3}^{2}+{3}^{2}}=3\sqrt{2}$，
$\therefore OE=CE-OC=3\sqrt{2}-3$.
故答案为：$3\sqrt{2}-3$.