Let $h,r$ be the height and the radius in centimeters of the bottle, respectively. The volume of the bottle is $\frac{h \cdot \pi r^{2}}{3}$. When the bottle lies on its base, the volume occupied by the water is obtained by subtracting the volume of the cone we get cutting the bottle with a plane parallel to its base $8 $ centimeters under the vertex to the volume of the bottle itself. That cone that we obtain is similar to the bottle and the ratio between their elements is $8:h$. Therefore the volume of the water, in this case, is given by $\frac{h \cdot \pi r^{2}}{3} - \frac{8 \cdot \pi (\frac 8h r)^{2}}{3}$ $[1]$ When the bottle is upside down, the volume of the water is just the volume of the cone we get cutting the bottle with a plane parallel to the base $2 $ centimeters above the vertex. Since that cone is similar to the bottle with a ratio $(h-2):h$, the volume of the water, in this case, is given by $\frac{h \cdot \pi (\frac{h-2}{h}r)^{2}}{3}$ $[2]$ Since the volume of the water doesn't change when it lies on its base or it's turned upside down, we consider the equation $[1]=[2]$. After some simplifications, we end up with $h^{2}-2h-84=0$, hence $h=1+\sqrt{85}$ $\blacksquare$