Answer
$v = 6.49~m/s$
Work Step by Step
We can use the energy $E$ to find the angular speed.
$\frac{1}{2}I\omega^2 = E$
$\frac{1}{2}(\frac{2}{5}mr^2)\omega^2 = E$
$\frac{1}{5}(mr^2)\omega^2 = E$
$\omega^2 = \frac{5E}{(mr^2)}$
$\omega = \sqrt{\frac{5E}{(mr^2)}}$
$\omega = \sqrt{\frac{(5)(236~J)}{(28.0~kg)(0.380~m)^2}}$
$\omega = 17.08~rad/s$
We can use the angular speed to find the tangential velocity.
$v = \omega ~r = (17.08~rad/s)(0.380~m)$
$v = 6.49~m/s$
