Let the height of the tower be $h$ and the common initial speed be $u$.

Take downward direction as positive, and acceleration due to gravity as $g$.

Equation for the first ball (thrown vertically upward):

Since upward is negative in our convention, initial velocity is $-u$. It reaches the ground in $6\,\text{s}$, so $h = -u(6) + \frac{1}{2}g(6)^2$ $h = -6u + 18g \quad \text{(1)}$

Equation for the second ball (thrown vertically downward):

Initial velocity is $+u$. It reaches the ground in $1.5\,\text{s}$, so $h = u(1.5) + \frac{1}{2}g(1.5)^2$ $h = 1.5u + 1.125g \quad \text{(2)}$

Equate (1) and (2) since both are equal to $h$:

$-6u + 18g = 1.5u + 1.125g$ $16.875g = 7.5u$ $u = \frac{16.875}{7.5}g = 2.25g$

Find the height $h$ using equation (2):

$h = 1.5(2.25g) + 1.125g$ $h = 3.375g + 1.125g = 4.5g$

For the third ball, it is released from rest, so initial velocity is $0$.

Hence, $h = \frac{1}{2}gt^2$

Substitute $h = 4.5g$: $4.5g = \frac{1}{2}gt^2$ $4.5 = \frac{1}{2}t^2$ $t^2 = 9$ $t = 3\,\text{s}$

Final answer:

The third ball reaches the ground in $\boxed{3\,\text{s}}$