Here, $a = 2\, mm = 2 \times 10^{- 3}\, m$
$\lambda = 500\, nm = 500 \times 10^{-9}\, m = 5 \times 10^{-7}\, m, D = 1\,m$
The distance between the first minima on either side on a screen
$ = \frac{2\lambda D}{a} = \frac{2\times5\times10^{-7}\times1}{2\times10^{-3}}$
$= 5 \times10^{-4} m $
$ =0.5\times10^{-3} m$
$ = 0.5 \,mm$