If either cone is flattened the resultant figure is a circle with a sector cut out of it. The circle has a radius equal to the slant length L of the cone. If the sector has an angle θ, then the lateral area of the cone is πL²-½L²θ, being the difference between the area of the circle and the area of the sector.
The circumference of this area is 2πL-Lθ, being the circumference of the circle less the length of the arc of the sector. This circumference is the circumference of the base of the cone, 2πR where R is the radius of the base. So 2πR=2πL-Lθ, from which L(2π-θ)=2πR, and L/R=2π/(2π-θ). For the larger cone L/R=15/9=5/3.
Therefore 5/3=2π/(2π-θ), 10π-5θ=6π, θ=⅘π. This applies to both cones because they are similar. We know the lateral area (surface area) of the smaller cone is 60π=πL²-⅖πL²=⅗πL², therefore L²=100, L=10. So, since L/R=5/3, R=6. The height H of the smaller cone is √(10²-6²)=8 so the volume is ⅓R²H=⅓×36×8=96cc (cubic centimetres).