jmc

We start by finding the radius of the inscribed sphere. If we slice the diagram by a plane that contains the central axis of the cone, we get a circle inscribed in an isosceles triangle with base 6 and height 4, and the radius of the inscribed circle is the same as the radius of the sphere (since any plane that contains the central axis of the cone contains a diameter of the inscribed sphere). We label the points as shown in the diagram below.



Since $AD$ has length 3 and $DB$ has length 4, segment $AB$ has length 5, from the Pythagorean theorem. Similarly, segment $CB$ has length 5. Now, the area of triangle $ABC$ is equal to the semiperimeter times the radius of the inscribed circle. On the other hand, we know that the area of $ABC$ is also $$\frac{1}{2}AC\cdot DB=\frac{1}{2}\cdot 6\cdot 4=24/2.$$Let $\rho$ be the radius of the inscribed circle, and let $s$ be the semiperimeter of $ABC$. We then have $$\frac{24}{2}=\rho s=\rho \cdot \frac{AB+BC+AC}{2}=\rho \cdot \frac{16}{2}.$$Therefore $$\rho =\frac{24}{16}=3/2.$$Thus the volume of the inscribed sphere is $\frac{4}{3}\pi {\rho }^3=\frac{4}{3}\pi (3/2{)}^3$.

On the other hand, the volume of a cone with radius $r$ and height $h$ is $\frac{\pi }{3}{r}^{2}h$, so the volume of our cone is $$\frac{\pi }{3}\cdot 3^2\cdot 4.$$Therefore the ratio of the volume of the sphere to the volume of the cone is $$\frac{(4\pi /3)(3/2{)}^3}{(\pi /3)\cdot 3^2\cdot 4}=\frac{4\cdot 27/8}{9\cdot 4}=\boxed{\frac{3}{8}}.$$