As the body is dropped from a certain height $h$, its initial velocity will be zero.

For first half, $h_1=\frac{h}{2}$.

Using second equation of motion, $s=ut+\frac{1}{2}a{t}^2$

We have, $\frac{h}{2}=\frac{1}{2}g{t}_1^2$.

Time taken in travel first half of the height, $t_1=\sqrt{\frac{h}{g}} ...\left(1\right)$.

For total height $h$, using again second equation of motion,

$h=\frac{1}{2}g{\left(t_1+t_2\right)}^2\Rightarrow \left(t_1+t_2\right)=\sqrt{\frac{2h}{g}} ...\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, 

$\Rightarrow \frac{1}{\sqrt{2}}=\frac{t_1}{t_1+t_2}$

$\Rightarrow 1+\frac{t_2}{t_1}=\sqrt{2}$

$\Rightarrow \frac{t_1}{t_2}=\frac{1}{\sqrt{2}-1}$

Thus, the relation is $t_2=\left(\sqrt{2}-1\right)t_1$.