Answer
$-80 \pi$
Work Step by Step
The flux through a surface can be defined only when the surface is orientable.
We know that $\iint_S F \cdot dS=\iint_S F \cdot n dS$
Here, $n$ denotes the unit vector.
Since, $\iint_S f(x,y,z) dS =g(\sqrt {x^2+y^2+z^2})=g(\sqrt 4) =g(2)=-5$
Thus, $\iint_S f(x,y,z) dS =-5 \iint_S dS$
Here, $\iint_S dS$ is the surface area of the sphere $=4 \pi r^2=4 \pi (4)=16 \pi$
Thus, $\iint_S f(x,y,z) dS =-5 (16 \pi)=-80 \pi$
