When a dielectric is inserted into an isolated and charged capacitor,…

The correct answer is Option (3) → 3

for an isolated capacitor, (Q = Constant)

$U =\frac{1}{2}\frac{Q^2}{C}$

When a dielectric is inserted,

$C'=KC$

So the new energy,

$U' =\frac{1}{2}\frac{Q^2}{KC}=\frac{1}{K}\left(\frac{1}{2}\frac{Q^2}{C}\right)$

$=\frac{U_0}{K}$

$U'=33\%.U_0=\frac{33}{100}U_0$

$⇒\frac{U_0}{K}=\frac{33}{100}U_0⇒K=\frac{100}{33}≃3$