S base circle = π r 2,

Column height h=2R,

V=πR^2*h=2πR^3，

r=[v/(2π)]^( 1/3),( 1)

At the bottom, let the side length of a regular triangle be a and the height of the triangle be √3a/2. According to the nature of the center of gravity,

R=(2/3)*(√3a/2)=√3a/3, (R is the center line of 2/3, and the height and center line are integrated),

Then a=√3R,

Bottom area s △ ABC = √ 3a2/4 = √ 3 * (3R2)/4 = 3 √ 3R2/4,

V triangular prism = s △ ABC * h = (3 √ 3r 2/4) * 2r = 3 √ 3r 3/2,

Replace with (1),

∴V prism =3√3V/(4π).