A sphere in a quarter cone

Using cylinder coordinate:

By studying the position of sphere in the quarter cone, we know that the sphere will tangent to two side of vertical plane and quarter circle at the bottom.

So, the coordinate of center of sphere will be $(r,r,r)$ .

Let see plane $\theta = \frac{\pi}{4}$

Sphere will tangent to the incline line at this position.

So distance of center of sphere to the incline line will be radius of sphere,

or we can do it equivalently, from center of sphere to the plane $\frac{x}{\sqrt2 R}+\frac{y}{\sqrt2 R}+\frac{z}{H}=1$

$r=|\frac{\frac{\sqrt2 r}{R}+\frac{r}{H}-1}{\sqrt{\frac{1}{R^2}+\frac{1}{H^2}}}|$

$r=|\frac{r(\sqrt2 +\frac{R}{H})-R}{\sqrt{1+\frac{R^2}{H^2}}}|$

Hence, we can compute $\frac{H}{R}=2$ now,

Doing some work, found that

$(\sqrt2+1)(\frac{r}{R})^2-(2\sqrt2 +1)(\frac{r}{R})+1=0$

Solving this to get $\frac{r}{R}=1.256 or 0.3298$

but since the ratio giving no sense if it bigger than 1,

So $\frac{r}{R}=\boxed{0.330}$