Question 903386
The volume of a spherical cap of height {{{h}}} is 
{{{V[cap]=(pi/3)h^2(3R-h)}}}
So the volume of the remaining portion of the sphere is,
{{{V[left]=(4/3)pi*R^3-(pi/3)h^2(3R-h)}}}
Since it's a unit sphere, {{{R=1/2}}}
and
{{{V[left]=3V[cap]}}}
Substituting,
{{{(4/3)pi(1/2)^3-(pi/3)h^2(3(1/2)-h)=3(pi/3)h^2(3(1/2)-h)}}}
{{{(4/3)pi(1/8)=4(pi/3)h^2(3/2-h)}}}
{{{h^2(3/2-h)=1/8}}}
Here is a graph of {{{h^2(3/2-h)-1/8=0}}}
{{{graph(300,300,-2,2,-2,2,x^2(3/2-x)-1/8)}}} with x as h.
There are three solutions.
One is negative.
One is greater than 1.
The remaining solution using Newton's method to solve (after 5 iterations)
{{{h=0.3263518223}}}
Since this is the height of the cap, the distance from the center of the sphere is,
{{{x=R-h}}}
{{{x=1-0.3263518223}}}
{{{x=0.6736481777}}}