Notice in the given figure, consider right triangles
The normal height of the smaller cone (cap of the frustum) $$=r\cot\alpha$$ Volume of smaller cone with circular base of radius $r$ & normal height $r\cot\alpha$ $$V_1=\frac{1}{3}\pi r^2(r\cot\alpha)=\frac{1}{3}\pi r^3\cot\alpha$$
The normal height of the larger cone (frustum with its cap) $$=R\cot\alpha$$ Volume of larger cone with circular base of radius $R$ & normal height $R\cot\alpha$ $$V_2=\frac{1}{3}\pi R^2(R\cot\alpha)=\frac{1}{3}\pi R^3\cot\alpha$$
Now, the volume of the frustum of cone $$V=\text{(volume of larger cone)} - \text{(volume of smaller cone)}$$ $$V=V_1-V_2$$ $$V=\frac{1}{3}\pi R^3\cot\alpha-\frac{1}{3}\pi r^3\cot\alpha$$ $$R^3=\frac{3}{\pi\cot\alpha}\left(V+\frac{1}{3}\pi r^3\cot\alpha\right)=\frac{3V}{\pi\cot \alpha}+r^3$$ $$\color{red}{R=\left(\frac{3V}{\pi\cot\alpha}+r^3\right)^{1/3}}$$