If ball A is thrown up with half the speed of B, it will not go as high, therefore will fall to the ground quicker. 
Using the suvat equation $s=v_{0}t+\frac{1}{2}at^2$

For A, calculate when $s=0$ to find when the ball reached the ground
$$0=v_{0}t-4.5t^2$$
Solve for t to get $t=0$ and $t=\frac{v_{0}}{4.5}$

Do the same for B:
$$0=2v_{0}t-4.5t^2$$
$t=0$ and $t=\frac{2v_{0}}{4.5}$

This shows that it takes twice as long for B to reach the ground