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
$$0=v_{0}t-4.5t_A^2$$
Solve for $t_A$ to get $t_A=0$ and $t_A=\frac{v_{0}}{4.5}$

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

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

To find the heights reached by each ball use the equation $v^2=v_0^2+2as$. For A this will give:
$$0=v_0^2-19.6s_A$$
$$s_A=\frac{v_0^2}{19.6}$$
The same can be done for B to give $s_B=\frac{4v_0^2}{19.6}$.This shows that B will reach a height 4 times greater than A.