Explanation

(d) The potential at a distance $r$, due to charge $q$ is
$V = \frac{1}{4\pi\varepsilon_{0}}.\frac{q}{r}$
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Potential at a distance (3r) is
$V = \frac{1}{4\pi\varepsilon_{0}}.\frac{q}{3r}$
Difference in potential
$= \frac{q}{4\pi\varepsilon_{0}}\left\lbrack \frac{1}{r} - \frac{1}{3r} \right\rbrack$
$\Longrightarrow V = \frac{2q}{4\pi\varepsilon_{0} \times 3r}$
Intensity of electric field
$E = \frac{q}{4\pi\varepsilon_{0}}\frac{1}{(3r)^{2}}$
$\therefore\frac{E}{V} = \frac{q}{4\pi\varepsilon_{0}qr^{2}} \times \frac{4\pi\varepsilon_{0}3r}{2q}$
$\Longrightarrow \frac{E}{V} = \frac{1}{6r}\ $
$\Longrightarrow E = \frac{V}{6r}$