（5分）已知向量$\overrightarrow{a}$，$\overrightarrow{b}$满足$\vert \overrightarrow{a}-\overrightarrow{b}\vert =\sqrt{3}$，$\vert \overrightarrow{a}+\overrightarrow{b}\vert =\vert 2\overrightarrow{a}-\overrightarrow{b}\vert$，则$\vert \overrightarrow{b}\vert =$____．
分析：根据向量数量积的性质及方程思想，即可求解．
解：$\because \vert \overrightarrow{a}-\overrightarrow{b}\vert =\sqrt{3}$，$\vert \overrightarrow{a}+\overrightarrow{b}\vert =\vert 2\overrightarrow{a}-\overrightarrow{b}\vert$，
$\therefore$${\overrightarrow{a}}^{2}+{\overrightarrow{b}}^{2}-2\overrightarrow{a}\cdot \overrightarrow{b}=3$，${\overrightarrow{a}}^{2}+{\overrightarrow{b}}^{2}+2\overrightarrow{a}\cdot \overrightarrow{b}=4{\overrightarrow{a}}^{2}+{\overrightarrow{b}}^{2}-4\overrightarrow{a}\cdot \overrightarrow{b}$，
$\therefore$${\overrightarrow{a}}^{2}=2\overrightarrow{a}\cdot \overrightarrow{b}$，$\therefore$${\overrightarrow{b}}^{2}=3$，
$\therefore$$\vert \overrightarrow{b}\vert =\sqrt{3}$．
故答案为：$\sqrt{3}$．
点评：本题考查向量数量积的性质及方程思想，属基础题．