Position of $n$ th bright fringe from central maxima
$x_{n_{1}}=\frac{n_{1} \lambda D}{d}$ here $n_{1}=5$
$\therefore x_{n_{1}}=\frac{5 \lambda D}{d}$
Position of $n$ th dark fringe from central maxima
$x_{n} =\frac{(2 n-1) \lambda D}{2 d},$ Here $n=3$
$x_{n} =\frac{5}{2} \frac{\lambda D}{d}$
$x_{n_{1}}-x_{n} =\frac{2.5 \lambda D}{d}=2.5 \beta$
Given $\beta=0.4 mm$
$\Rightarrow x_{n_{1}}-x_{n} =1 mm$