Answer
$0.23^{\circ}$
Work Step by Step
We can use the values given with equation (36-3) to find the angle, $\theta$;
$\alpha \sin \theta = m \lambda $ where $m=1,2,3,....$ (minima)
Solving for $\theta$, we have:
$ \theta =\arcsin( m \lambda /\alpha )/2$
$ \theta =\arcsin( 1 *0.010m /(5.0m ))$
$ \theta =0.11459^{\circ}$
Because they want the angular width of the central diffraction maximum, we must double the angle, $\theta$ to obtain :
$2 \theta =2*0.11459^{\circ}\approx0.23^{\circ}$
