Answer
$3.1\times10^{-16} \ Js$
Work Step by Step
We first find the mass of the disk:
$M = \pi(150\times10^{-6})^2(2\times10^{-6})(2329)=3.293\times10^{-10}$
Thus, we find:
$L=I\omega=\frac{1}{2}MR^2\omega=\frac{1}{2}(3.293\times10^{-10})(150\times10^{-6})^2(83.77)=3.1\times10^{-16} \ Js$
