解：设该圆锥底面圆的半径为$r\ cm.$ 过点$O$作$OD\perp AB，$垂足为$D.$
$\because OA = OB，$$\angle AOB = 120^{\circ}，$$\therefore AD = BD=\frac{1}{2}AB = \sqrt{3}\ cm，$
$\angle OAD=\frac{1}{2}\times(180^{\circ}-120^{\circ}) = 30^{\circ}.$$\therefore OD=\frac{1}{2}OA.$
由勾股定理，得$AD^{2}+OD^{2}=OA^{2}，$即$(\sqrt{3})^{2}+(\frac{1}{2}OA)^{2}=OA^{2}.$$\therefore OA = 2\ cm.$
$\therefore 2\pi r=\frac{120\pi\times2}{180}，$解得$r = \frac{2}{3}.$$\therefore$该圆锥底面圆的半径为$\frac{2}{3}\ cm.$
