The centripetal acceleration must be
$\begin{align*}a = \frac{v^2}{r} = \omega^2 r\end{align*}$
So, $m \omega^2 r$ must be the horizontal component of the tension in the cord. The mass is not accelerating in the vertical direction, so the vertical component of the tension must be $mg$ . So, the tension in the cord is:
$\begin{align*}F_T = \sqrt{(m g)^2 + (m \omega^2 r)^2 } = \boxed{m \sqrt{g^2 + \omega^4 r^2}}\end{align*}$
Therefore, answer (E) is correct.

Solution to 1986 Problem 37