Let the uniform angular acceleration be $\alpha$. Given initial angular velocity, $\omega_0 = 0$. Angle rotated in the first $2\text{ s}$ is $\theta_1 = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + \frac{1}{2}\alpha (2)^2 = 2\alpha$. Angle rotated in the first $4\text{ s}$ (i.e., $2\text{ s} + 2\text{ s}$) is $\theta_{total} = \omega_0 t' + \frac{1}{2}\alpha t'^2 = 0 + \frac{1}{2}\alpha (4)^2 = 8\alpha$. The additional angle rotated in the next $2\text{ s}$ is $\theta_2 = \theta_{total} - \theta_1 = 8\alpha - 2\alpha = 6\alpha$. Therefore, the ratio $\frac{\theta_2}{\theta_1} = \frac{6\alpha}{2\alpha} = 3$.