For central maxima, \(\sin \theta=\frac{\lambda}{a}\)
Also, \(\theta\) is very-very small so
\(\sin \theta \approx \tan \theta=\frac{y}{D}\)
\(\therefore \quad \frac{y}{D}=\frac{\lambda}{a}, y=\frac{\lambda D}{a}\)
Width of central maxima \(=2 y=\frac{2 \lambda D}{a}\)