The lines connecting the center of marbles will form into a

regular tetrahedron of edge 20 cm.

From triangle ABC

$\cos 30^\circ = \dfrac{10}{AE}$

$AE = \dfrac{10}{\cos 30^\circ} = \dfrac{10}{\frac{\sqrt{3}}{2}}$

$AE = 11.55 \, \text{ mm}$

$CE = AE$

$CE = 11.55 \, \text{ mm}$

Radius of cylinder

$R = 10 + AE = 10 + 11.55$

$R = 21.55 \, \text{ mm}$

Solving for h from right triangle CED

$h^2 + CE^2 = CD^2$

$h^2 + 11.55^2 = 20^2$

$h = 16.33 \, \text{ mm}$

Depth of water

$H = 20 + h = 20 + 16.33$

$H = 36.33 \, \text{ mm}$

Volume of water

$V_{water} = V_{cylinder} - 4V_{marble}$

$V_{water} = \pi (21.55^2)(36.33) - 4 [ \, \frac{4}{3}\pi (10^3) \, ]$

$V_{water} = 36\,248.98 \, \text{ mm}^3$ *answer*