Put $nx = t$ , $ndx = dt$ and $x = t/n$ we get
$\lim_{n \rightarrow \infty }n \int_{0}^{100} f(x) \psi (nx) dx = \lim_{n \rightarrow \infty }\int_{0}^{100n} f(t/n) \psi (t) dt$
$ = \int_{0}^{1} f(0) \psi (t)dt$ .
The last step is really important $t/n$ approaches 0 . $100n>>1$ and we dont have to go beyond 1 because $\psi (y) = 0$ elsewhere as given in question.
Continuing on $ = f(0) \int_{0}^{1} \psi (t)dt = f(0)* 1 = f(0)$ .
So $A$ is correct.