Take upward as positive.

Initial velocity of the ball from the top of the tower: $u=19.6\,\text{m/s}$ Acceleration due to gravity: $a=-9.8\,\text{m/s}^2$

Use the given total time to hit the ground.

Let the height of the tower be $h$.

Position after time $t$ from the point of projection is $y=ut+\frac12 at^2$

When the ball hits the ground after $6$ s, its displacement from the top of the tower is $-h$: $-h=19.6(6)+\frac12(-9.8)(6)^2$

$-h=117.6-4.9\times 36$

$-h=117.6-176.4=-58.8$

Hence, $h=58.8\,\text{m}$

Find how high the ball rises above the point of projection.

At the highest point, $v=0$

Using $v^2=u^2-2gs$ $0=(19.6)^2-2(9.8)s$

$s=\frac{(19.6)^2}{19.6}=19.6\,\text{m}$

So the ball rises $19.6$ m above the top of the tower.

Maximum height above the ground

$H=h+s=58.8+19.6=78.4\,\text{m}$

Match with the form $\left(\frac{k}{5}\right)$ m

$\frac{k}{5}=78.4$

$k=78.4\times 5=392$

Final answer

$\boxed{k=392}$