Let the body reach the velocity $v$ in time $t$. Since the body is dropped, initial velocity is zero $(u=0).$
\[v=u+gt\] \[\therefore v=gt\]
If the new time is twice the original time, $t’=2t.$
Let the new velocity reached be $v’.$ \[v’=gt’=g(2t)=2(gt)=2v\]
Thus, the new velocity will be twice the original velocity.
[Equations used above are deduced from Motion Under Gravity]