如图，设$AO$与$BC$交于点$D$.

$\because \angle AOB=60^{\circ}$，$AB=$$AB$，

$\therefore \angle C=\dfrac{1}{2}\angle AOB=30^{\circ}$，

又$\because AB=AC$，

$\therefore$$AB=$$AC$

$\therefore AD\bot BC$，

$\therefore BD=CD$，

$\therefore$在直角$\triangle ACD$中，$CD=AC\cdot \cos 30^{\circ}=2\times \dfrac{\sqrt{3}}{2}=\sqrt{3}$，

$\therefore BC=2CD=2\sqrt{3}$.

故选：$C$.