Spherical Shelter

The left-hand picture is a top view. The cube must be small enough that its corners don't stick out. Let the sides of the cube be $2x$ , then

$AO=1, AB=x, BO=\sqrt{1-x^{2}}$

$CO=\sqrt{1-x^{2}} - 2x$

The right-hand picture is a side view

$O'D=\sqrt{1-x^{2}} - 2x, C'D=1-2x, O'C'=1$

So by the Pythagorean theorem

$(\sqrt{1-x^{2}})^{2}+(1-2x)^{2}=1$

This can be simplified to the quartic

$65x^{4}-56x^{3}+14x^{2}-8x+1=0$

The closed form is immense. The approximate real root is $x \approx .14385584$

$\boxed{2x \approx .2877}$

Note that the answer to the 2D problem is $\frac{2-\sqrt{2}}{2}$ .

Hint: Use the pythagorean theorem to show that $\sqrt{2}a$ is the length of a diagonal in a square with side length $a$ and apply this fact twice to get the answer.