See the diagram.

let the solid ball have a radius of R cm.

From the diagram: CD = 8 cm and AB = 24 cm given.

Hence, DB = 24/2 = 12 cm.

From the triangle ODB, we get: OD² + DB² = OB² -- Pythagoras theorem.

=> (R-8)² + 12² = R²

=> 16 R = 208 * => R = 13 cm*

Radius of the ball = 13 cm.

=============

*we can also find the relative density of the ball.*

Let the ball have a density of d gm/cm³.

mass of the ball = 4/3 π 13³ * d = 9,202.78 * d gms

*Volume of* the ball immersed in water is found using integral calculus with limits of y = OD to OC. Let h = OD = 5 cm.

Volume of water displaced = 2 π R³ / 3 - π h [ R² - h² / 3 ]

= 2077.64 cm³

weight of water displaced = volume * density = 2077.64 cm³ * 1 gm/cm³ = 2077.64 gms

weight of the ball = weight of the water displaced , Archimedes principle.

2077.64 = 9, 202.78 * d

* d = density of the ball = 0.226 gm/cm³*