#### Answer

$ .824\ Nm$

#### Work Step by Step

We first must find the moment of inertia:
$ I = \frac{2}{3}mr^2 = \frac{2}{3}(.622)(.12^2)=.006$
Thus, we now can find the value of k by simplifying the equation for the angular velocity of a wire and by recalling that omega is equal to $2\pi f$. Doing this, we find:
$k = 4\pi^2f^2I = 4\pi^2(1.87^2)( .006) = .824\ Nm$