$\because$分别以$AC$，$AB$为边向外作正方形，面积分别为$S_{1}$，$S_{2}$，
$\therefore {S}_{1}=A{C}^{2}$，${S}_{2}=A{B}^{2}$.
$\because S_{1}=3$，$S_{2}=7$，
$\therefore AC^{2}=3$，$AB^{2}=7$.
$\because \angle ACB=90^{\circ}$，
$\therefore BC=\sqrt{A{B}^{2}-A{C}^{2}}=\sqrt{7-3}=2$.
故答案为：$2$.

$\because$分别以$AC$，$AB$为边向外作正方形，面积分别为$S_{1}$，$S_{2}$，
$\therefore {S}_{1}=A{C}^{2}$，${S}_{2}=A{B}^{2}$.
$\because S_{1}=3$，$S_{2}=7$，
$\therefore AC^{2}=3$，$AB^{2}=7$.
$\because \angle ACB=90^{\circ}$，
$\therefore BC=\sqrt{A{B}^{2}-A{C}^{2}}=\sqrt{7-3}=2$.
故答案为：$2$.