Answer
$(x-3)^2+(y-\frac{1}{2})^2+(z+1)^2=16$
Center: $(3,\frac{1}{2},-1)$ and Radius: $4$
Work Step by Step
$4x^2+4y^2+4z^2-24x-4y+8z-23=0$, and so dividing by $4$
$x^2+y^2+z^2-6x-y+2z-\frac{23}{4}=0$
$=(x-3)^2-9+(y-\frac{1}{2})^2-\frac{1}{4}+(z+1)^2-1-\frac{23}{4}$
$=(x-3)^2+(y-\frac{1}{2})^2+(z+1)^2-16=0$
$(x-3)^2+(y-\frac{1}{2})^2+(z+1)^2=16$
Center: $(3,\frac{1}{2},-1)$ and Radius: $4$