Answer
$\omega_B=211 rad/s$
Work Step by Step
We can determine the required angular velocity as follows:
$\int_0^{\omega_A}d\omega=\int_0^t \alpha dt$
$\int_0^{\omega_A}d\omega=\int_0^{1.5} 400 t^3 dt$
This simplifies to:
$\omega_A=(\frac{400}{4}t^4)|_0^{1.5}$
$\implies \omega_A=506.25 rad/s$
As gear A and gear B are meshed together, therefore
$\omega_A r_A=\omega_B r_B$
This can be rearranged as:
$\omega_B=\frac{r_A}{r_B}\omega_A$
We plug in the known values to obtain:
$\omega_B=(\frac{0.5}{1.5})(506.25)$
$\implies \omega_B=211 rad/s$
