Finding the largest cube inscribed in a hemisphere has been considered here previously. So let's consider the reverse relationship:
A hemisphere is inscribed in a cube with an edge of $1m$. What is the maximum radius of the hemisphere ?
Obviously a whole sphere of radius $\frac{1}{2}$ can be inscribed in the cube, but could a hemisphere of larger radius fit in there somehow?
On
Extending my comment on Win Vineeth's answer:
Imagine a flat circular disk $D$ of radius $1$ centered on the line $t\mapsto(t,t,t)$, and lying in a plane orthogonal to this line, such that it just touches the coordinate planes in three points forming an equilateral triangle.
Doing a little coordinate geometry one computes $$M=\sqrt{2\over3}(1,1,1)\ ,$$ and the $(x,y)$-plane is touched at $\bigl(\sqrt{3\over2},\sqrt{3\over2},0\bigr)$. Erect a hemiball of radius $1$ on the "far side" of $D$. This hemiball will then fit into the cube $$\left[0,\sqrt{2\over3}+1\right]^3\ .$$ If the cube is required to have side length $1$ the admissible radius $\rho$ of the hemiball is therefore given by $$\rho={1\over \sqrt{2/3}+1}=3-\sqrt{6}\doteq0.5505\ .$$
Maximum radius without any doubt is $\frac 12m$
Imagine the hemisphere with the base on a side of the cube.