解：过点$O$作$OC\perp AB$于点$C，$则$AC = \frac{1}{2}AB，$$\angle OCA = 90^{\circ}.$ 因为圆心$O$到栏杆$AB$的距离是$5\ m，$所以$OC = 5\ m.$ 因为$\odot O$的半径是$10\ m，$所以$OA = 10\ m，$所以$AC = \sqrt{OA^{2}-OC^{2}} = 5\sqrt{3}\ m，$所以$AB = 2AC = 10\sqrt{3}\ m，$所以$S_{\triangle OAB} = \frac{1}{2}AB\cdot OC = 25\sqrt{3}\ m^{2}.$ 因为$OC = \frac{1}{2}OA，$所以$\angle OAB = 30^{\circ}.$ 因为$OA = OB，$所以$\angle OBA = \angle OAB = 30^{\circ}，$所以$\angle AOB = 180^{\circ}-\angle OAB - \angle OBA = 120^{\circ}，$所以$S_{扇形OAB} = \frac{120\pi\times10^{2}}{360} = \frac{100\pi}{3}(m^{2})，$所以$S_{阴影} = S_{扇形OAB}-S_{\triangle OAB} = (\frac{100\pi}{3}-25\sqrt{3})m^{2}\approx 61.4\ m^{2}.$ 因为$3\times61.4\approx 184$(名)，所以该场馆最多可容纳$184$名观众同时观看演出.
