The diameter of the sphere is equal to the length of the diagonal between opposite corners in the cube.
Let [tex]x[/tex] be the length of this diagonal, and let [tex]y[/tex] be the length of the diagonal on any face of the cube. By the Pythagorean theorem,
[tex]3^2 + 3^2 = y^2 \implies y = \sqrt{18} = 3\sqrt2[/tex]
[tex]3^2 + y^2 = x^2 \implies x = \sqrt{27} = 3\sqrt3[/tex]
(see the attached sketch (not mine))
Then the radius of the sphere is half of this,
[tex]\boxed{\dfrac{3\sqrt3}2}[/tex]
