The volume is given by
[tex]V=\pi hr^2[/tex]
Since the height is 3 times the radius of the base, we can substitute [tex]h=3r[/tex]:
[tex]V=\pi (3r)r^2=3\pi r^3[/tex]
And we know that the volume is [tex]24\pi[/tex] (I'm actually assuming that there's a typo in the question, otherwise the options wouldn't match):
[tex]24\pi=3\pi r^3 \iff r^3=8 \iff r=2[/tex]
The height is 3 times the radius, so it is
[tex]h=2\cdot 3 = 6[/tex]
