The volume of water in the vessel of the base area $A$ and height $h$ is$V=Ah$. The average velocity of outflowing water when the height of water changes from $h$ to 0 is

$v=\frac{\sqrt{2 \mathrm{g}h}+0}{2}=\frac{\sqrt{2\mathrm{g}h}}{2}$

$\therefore V=A v t$ ...(i)

When the vessel is filled to height 4 $h$, then the volume in the vessel

$=4V=4A vt=4A\frac{\sqrt{2\mathrm{g}h}}{2}\times t$

If $t$ is the time taken for the outflowing liquid and $v_1$ is the average velocity of outflowing liquid then

$4V=A v_1{t}_1$

or $t_1=\frac{4 V}{A{ v}_1}=\frac{4 A\sqrt{2\mathrm{g}h}\times t\times 2}{2\times A\times \sqrt{2\mathrm{g}\times 4h}}=2t$