Let $$V = \left\{ (x, y, z)\in \mathbb R^3 : \tfrac{1}{4}\le x^2+y^2+z^2\le 1\right\}$$ and $$f = \frac{xi+yj+zk}{(x^2+y^2+z^2)^2} \text{ for } (x, y, z) \in V.$$ Let $n$ denote an outward unit normal vector to the boundary of $V$ and $S$ denote the part $\{ (x, y, z) \in \mathbb R^3: x^2+y^2+z^2=1/4\}$ of the boundary of $V$. Then calculate $$\iint_s f\cdot n\,dS.$$
In this solution, the highlighted area look problem to me, the book has substituted the $x^2+y^2+z^2$ in the denominator by $1/4$, using the equation of of surface S. But is that allowed, since $1/(x^2+y^2+z^2)^{3/2}$ is the integrand which should be evaluated over surface $S$.
It is "allowed", yes. This stuff used to drive me nuts, so I know exactly where your head is right now, at least I think I do. I used to think of integrals very algorithmically / algebraically, but in fact it's just adding up a bunch of numbers multiplied by tiny numbers ( the dS ), and taking limits. So in this case, the numbers being added up are function $\frac{1}{(x^2+y^2+z^2)^{3/2}}$ evaluated at a point of the surface. But that function is always $1/4$ on the surface, so you don't really need to work very hard to evaluate it.