Why would you throw like that?

Lets first calculate the time the stone takes to reach up.
$\begin{aligned} \text{v} & = \text{u + at}\\ 0 & = 20 - 10\text{t}\\ -10\text{t} & = -20\\ \text{t} & = 2 \end{aligned}$

So, the stone takes the same time to reach from the top to the level of the building's top(where I was standing).

Now, let us calculate the time it takes to reach from the top of the building to the ground.
$\begin{aligned} \text{S} & = \text{ut} + \dfrac{1}{2}{\text{at}}^2\\ \\ 80 & = 20\text{t} + \dfrac{1}{2}×10+{\text{t}}^2\\ \\ 80 & = 20\text{t} + 5{\text{t}}^2\\ \\ 5{\text{t}}^2 + 20\text{t} - 80 & = 0\\ \\ {\text{t}}^2 + 4\text{t} - 16 & = 0\\ \\ \text{Using quadratic formula}:-\\ \\ \text{t} & = \dfrac{-4 \pm \sqrt{4^2 - 4×1×-16}}{2×1}\\ \\ \text{t} & = \dfrac{-4 \pm \sqrt{80}}{2}\\ \\ \text{t} & = \dfrac{-4 + 4\sqrt{5}}{2} \left(\text{or}\right) \dfrac{-4 - 4\sqrt{5}}{2}\\ \\ \text{But, time cannot be negative, so}:-\\ \\ \text{t} & = -2 + 2\sqrt{5}\\ \\ \text{So, total time} & = -2 + 2\sqrt{5} + 2 + 2\\ \\ & = 2\sqrt{5} + 2\\ \\ & = 2\left(\sqrt{5} + 1\right)\\ \\ & = 2× 3.23\\ \\ & \approx \color{#3D99F6}{\boxed{6.47}} \end{aligned}$

We can directly apply the $2^{\text{nd}}$ equation of motion,

$s=ut+\frac{1}{2}at^2 \\ 80=-20t+5t^2 \\ t=2±2\sqrt{5}$

I took vertically downward direction as $+\text{ve}$ and upward as $-\text{ve}$ .

As time can't be negative, $t=2+2\sqrt{5}\sim \boxed{6.47}$