jmc

Since the vertex angle of the cross-section triangle measures $60^{\circ }$, the cross-section triangle is equilateral. Also, the cross-section of the sphere inscribed in the cone is a circle tangent to each of the triangle's sides. Call the vertices of the equilateral triangle $A$, $B$, and $C$, and let $O$ be the center of the circle and $D$ and $E$ the midpoints of segments $AB$ and $BC$, respectively. To find the radius of the circle, divide the 30-60-90 triangle $CDB$ into three smaller congruent 30-60-90 triangles as shown. Since the area of each of these triangles is smaller by a factor of $3$ than the area of triangle $CDB$, each corresponding side must be smaller by a factor of $\sqrt{3}$. Thus $OE=DB/\sqrt{3}=6$ inches. Therefore, the volume of the sphere is $V=\frac{4}{3}\pi (\text{radius}{)}^3=\frac{4}{3}\pi (\text{6 inches}{)}^3=\boxed{288\pi }$ cubic inches.