K

Step 1: The capacitance of a capacitor is given by the formula $$ C = \frac{\varepsilon A}{d} $$ where $\varepsilon$ is the permittivity of the medium, $A$ is the area of the plates, and $d$ is the distance between them.
Step 2: When a dielectric material with dielectric constant $K$ is inserted into the capacitor, the new capacitance becomes $$ C' = K \cdot C $$, where $C$ is the original capacitance.
Step 3: The energy stored in a capacitor is given by the formula $$ U = \frac{1}{2} C V^2 $$, where $V$ is the voltage across the capacitor.
Step 4: Since the capacitor is isolated, the charge $Q$ remains constant. With the dielectric, the new voltage $V'$ is given by $$ V' = \frac{Q}{C'} = \frac{Q}{K C} = \frac{V}{K} $$.
Step 5: Substituting $C'$ and $V'$ into the energy formula: $$ U' = \frac{1}{2} C' (V')^2 = \frac{1}{2} (K \cdot C) \left(\frac{V}{K}\right)^2 = \frac{1}{2} (K \cdot C) \cdot \frac{V^2}{K^2} = \frac{C V^2}{2 K} = K \cdot U $$ where $U$ is the initial energy.
Step 6: Therefore, the energy stored in the capacitor becomes $K$ times of its initial value when the air is replaced by a medium of dielectric constant $K$.
Therefore, the answer is K.