This problem is more on your knowlefge of the parabolic flight of the bullet where air resistance is ignored.
The bullet fierd straight up will reach its max then it will fall down. At the maximum height, the vertical velocity is zero. Then at the same level where it was fired, the vertical velocity is the same as the initial velocity but in opposite direction.
Here it will be -30m/s.
That is the same as the initial velocity of the second bullet that is fired straight down.
That means the second bullet will be ahead of the first bullet by the amount of time that the first bullet spent from cliff level to max height to cliff level again.
At the max height, the vertical velocity of the first bullet is zero,
v = Vo -gt
0 = 30 -gt
t = 30/g seconds.
When the bullet falls to the level it was fired, it takes the same t = 30/g seconds again.
So it takes the first bullet 2(30/g) seconds to be in the same level where the second bullet was fired.
Therefore, whatever is the height of the cliff from the ground below, the first bullet will always be 60/g seconds late than the second bullet.
Hence, if your g = 10 m/sec/sec, then the first bullet will be 60/10 = 6 seconds late in hitting the ground.
If your g = 9.8 m/sec/sec, it would be 60/9.8 = 6.12245 seconds late.