(a) Suppose neutral point $N$ lies at a distance $x$ from dipole of moment $p$ or at a distance $(25$-$x)$ from dipole of 64 $p$
At $N$ |E.F. due to dipole | =|E.F. due to dipole |
$\Rightarrow \frac{1}{4\pi\varepsilon_{0}}.\frac{2p}{x^{3}} = \frac{1}{4\pi\varepsilon_{0}}.\frac{2(64p)}{(25 - x)^{3}}$
$\Rightarrow \frac{1}{x^{3}} = \frac{64}{(25 - x)^{3}} \Rightarrow x = 5\ cm$