The rock has height [tex]y[/tex] at time [tex]t[/tex] according to
[tex]y=v_0t-\dfrac g2t^2[/tex]
where [tex]v_0[/tex] is the velocity with which it was thrown, and g = 9.8 m/s^2 is the acceleration due to gravity.
Complete the square to get
[tex]y=\dfrac{{v_0}^2}{2g}-\dfrac g2\left(t-\dfrac{v_0}g\right)^2[/tex]
which indicates a maximum height of [tex]\dfrac{{v_0}^2}{2g}[/tex] occurs when [tex]t=\dfrac{v_0}g[/tex]. We're told this time is 2.6 s after the rock is thrown:
[tex]2.6\,\mathrm s=\dfrac{v_0}{9.8\frac{\rm m}{\mathrm s^2}}\implies v_0=25.48\dfrac{\rm m}{\rm s}[/tex]
So when t = 1.6 s, the rock reaches the tower's height of
[tex]y=v_0(1.6\,\mathrm s)-\dfrac g2(1.6\,\mathrm s)^2\approx\boxed{28\,\mathrm m}[/tex]
