Answer
$0.025mm$
Work Step by Step
We can use the we found in part a ($\theta_R= 8.8*10^{-7}rad$) with the given focal length of $f=14m$ to find the radius, $r$, of the dark ring.
$r=f\sin \theta$
$r=14m\sin 8.8*10^{-7}rad$
$r=1.232*10^{-5}m$
The diameter of the dark ring is twice the radius.
$d=2r$
$d=2 *1.232*10^{-5}m$
$d=2.464 *10^{-5}m$
$d=0.025mm$
