From the problem, we know that the area of the figure to the right of chord $AB$ is $\frac{2}{3}$ of the area of $\odot O_{2}$.

Let $\angle A O_{1} B=\alpha, \odot O_{2}$ have a radius of $r$, and take any point $C$ on the shaded circumference of $\odot O_{2}$, different from points $A$ and $B$. Then,
$$\begin{array}{l} S_{\triangle A O_{2} B}+S_{\text {sector } O_{2} A C B}=\frac{2}{3} \pi r^{2} \\ \Rightarrow \frac{1}{2} r^{2} \sin \alpha+\frac{2 \pi-\alpha}{2 \pi} \pi r^{2}=\frac{2}{3} \pi r^{2} \\ \Rightarrow \alpha-\sin \alpha=\frac{2 \pi}{3} . \end{array}$$

Using a TI graphing calculator to solve the equation, we get $\alpha=2.6053$ (radians).