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Cube here will be inscribed in a cone as a square is in isosceles triangle.
Let the height of the cone be $h$
Radius=$\sqrt2 h$
Volume of cone=$\dfrac{1}{3}\pi r^2h$
                           =$\dfrac{2\sqrt 2}{3}\pi h^3$
Let the side of the cube be x,the top of the cone above it has the sign $(h-x)$ and radius $\dfrac{x}{2}$
Using properties of similar triangle $\dfrac { \dfrac { x }{ 2 }  }{ h-x } =\dfrac{\sqrt2 h}{h}$
                                                            $=\sqrt 2 x$
                                                             $=\dfrac { 2\sqrt { 2 } h }{ 2\sqrt { 2 } +1 } $
Volume of the cube=$\dfrac { 2\sqrt { 2 } h }{ 2\sqrt { 2 } +1 } $
Ratio of the volume of the cone to volume of the cube=$\dfrac { \dfrac { 2\sqrt { 2 }  }{ 3 } \pi h^{ 3 } }{ (\dfrac { 2\sqrt { 2 } h }{ 2\sqrt { 2 } +1 } )^ 3 } $
                                            $=\dfrac{\pi(2\sqrt { 2 } +1  )^ 3)}{24}$
                                            $=2.35\pi$