Let the body travel the height $h$ in time $t$. Since the body is dropped, initial velocity is zero $(u=0).$

\[h=ut+\frac{1}{2}gt^2\] \[\therefore h=\frac{1}{2}gt^2\]

If the new time is twice the original time, $t’=2t.$

Let $h’$ be the new height travelled by the body. \[h’=\frac{1}{2}gt’^2=\frac{1}{2}g(2t)^2=4×\frac{1}{2}gt^2=4h\]

Thus, the new height will be four times the original.

[Equations used above are deduced from Motion Under Gravity]