Let in time $d t$ level of water in tank decreases by $d h$

Equating volumes, $A d h=d t \sqrt{2 g h} \times a$

$\Rightarrow d t=\frac{A}{a \sqrt{2 g}} \frac{d h}{\sqrt{h}}$

Integrating above we get,

$\int_{0}^{t} d t=\frac{A}{a \sqrt{2 g}} \int_{H_{2}}^{H_{1}} \frac{d h}{\sqrt{h}}$

$\Rightarrow t=\left.\frac{A}{a \sqrt{2 g}} \frac{h^{1 / 2}}{1 / 2}\right|_{H_{2}} ^{H_{1}}=\frac{A}{a} \sqrt{2 / g}[\sqrt{H_{1}}-\sqrt{H_{2}}]$