Answer : (d) 1.5
Explanation :
$tan \theta =\large\frac{qE}{mg}=constant$
$\large\frac{k\;q^2}{d^2\;mg}=tan 30^{0}$
$tan 30^{0}=\large\frac{k\;q^2}{d^2\;mg}=\large\frac{k\;q^2}{1.5 V g\;d^2}\;\quad\;$ [V is volume of sphere] ---1
When suspended in liquid :
$tan 30^{0}=\large\frac{1}{4 \pi \in}\;\large\frac{q^2}{d^2}\;\large\frac{1}{(mg-\sigma \;V\;g)}$ -----2
$=\large\frac{1}{4 \pi \in_{0} K} \;\large\frac{q^2}{d^2 gV (1.5-0.5)}$
from1 and 2 we get
$\large\frac{kq^2}{d^2 g V K}=\large\frac{kq^2}{1.5 V g d^2}$
$K=1.5$
K -------> dielectric constant of liquid
$k=\large\frac{1}{4 \pi \in_{0}}$