Answer
$\omega = 27.0~rad/s$
Work Step by Step
Let $\omega$ be the angular velocity at the start of the interval.
We can find $\omega$:
$\theta = \omega~t+\frac{1}{2}\alpha t^2$
$\omega~t = \theta - \frac{1}{2}\alpha t^2$
$\omega = \frac{\theta}{t} - \frac{1}{2}\alpha t$
$\omega = \frac{90.0~rad}{3.00~s} - (\frac{1}{2})(2.00~rad/s^2)(3.00~s)$
$\omega = 27.0~rad/s$
