$ \text{ For central Maximum } sin\theta = \frac{\lambda}{d} $

$\text{ Angular width is }2\theta = \frac{2\lambda}{d}$

$\text{Linear width is} = 2D\theta = \frac{2\lambda D}{d}$

$\text{According to the question} \frac{2\lambda D}{d} = d $

$\Rightarrow D = \frac{d^2}{2\lambda}$