The radius $\rho$ of the inscribed circle of a triangle can be found using $\rho=\dfrac{A}{s}$, where $s$ is the semiperimeter.

After you write the ratio of volumes, the $\pi$ and powers of $r$ should cancel. If they don’t, re-check your expressions.

Because $h=2r$, pick $r=1$ so $h=2$. Then the slant height is $\ell=\sqrt{1^2+2^2}=\sqrt{5}$.

In Desmos, enter the inradius formula for the triangle:

rho = (r*h)/(r+l)

with r=1, h=2, and l=sqrt(5). (You can directly type rho = (1*2)/(1+sqrt(5)).)

Enter:

ratio = ((4/3)*pi*rho^3)/((1/3)*pi*r^2*h)

using r=1 and h=2. Desmos will give a decimal value for ratio.

In Desmos, also type each answer choice as a decimal (for example, 2*sqrt(5)-4) and see which one matches the ratio value.

Slice the cone through its axis. The cross section is an isosceles triangle with:

• base $2r$ (the diameter of the cone’s base),
• height $h=2r$,
• equal sides (slant heights) $\ell=\sqrt{r^2+h^2}=\sqrt{r^2+(2r)^2}=r\sqrt{5}$.

The inscribed sphere becomes the incircle of this triangle, so the sphere radius equals the triangle’s inradius $\rho$.

Triangle area:

Semiperimeter:

So the inradius (sphere radius) is

Sphere volume:

Cone volume:

Thus

Since $\rho/r=\frac{2}{1+\sqrt5}$,

Use $\frac{1}{1+\sqrt5}=\frac{\sqrt5-1}{4}$ (because $(1+\sqrt5)(\sqrt5-1)=4$):

Now

so

Therefore, the correct choice is $2\sqrt{5}-4$.