The volume of a cone is h*A/3 where A is the area of the base.

Let h1 be the 8 cm height of air near the tip of the cone.

Let h3 be the full height of the cone.

Let h2 be the h3-2 cm filled with water when the point is held downward.

To avoid complications, just consider the fact that the three cones involved are all similar so the volume is proportional to the cube of the height. The area of the base is proportional to the square of the height and thus the volume is proportional to the cube of the height.

8^3 = h3^3 - h2^3

h3 - h2 = 2

While these are arbitrary units as regard volume, the 8 and h3 and h2 are all in centimeters, so the solution of these simultatneous equations should give us the answer in centimeters.

Calling h3, c; and referring to h2 as b; asking WolframAlpha

solve c^3-b^3=8^3, c-b=2 for c, b

gives h3 = c = 1 + sqrt(85) ~= 10.2195444572929

      h2 = b = sqrt(85) - 1 ~=  8.21954445729289

      

Answer: about 10.2195444572929 cm      

    or exactly  1 + sqrt(85) cm.