Five identical air-filled capacitors, each of capacitance C , are...

Initially, the five identical capacitors form a balanced Wheatstone bridge. The central capacitor is ineffective, and the equivalent capacitance is $C_0 = \dfrac{C \times C}{C + C} + \dfrac{C \times C}{C + C} = \dfrac{C}{2} + \dfrac{C}{2} = C$.

When a dielectric of $K = 4$ is introduced into the two capacitors connected to terminal $A$, their capacitances become $4C$.
Let the arms connected to terminal $A$ be $C_1 = 4C$ and $C_2 = 4C$.
Let the arms connected to terminal $B$ be $C_3 = C$ and $C_4 = C$.

The ratio of the adjacent arms is $\dfrac{C_1}{C_3} = \dfrac{4C}{C} = 4$ and $\dfrac{C_2}{C_4} = \dfrac{4C}{C} = 4$.
Since $\dfrac{C_1}{C_3} = \dfrac{C_2}{C_4}$, the bridge remains balanced. The central capacitor can still be ignored.

The new equivalent capacitance $C_{new}$ consists of two parallel branches, each having a $4C$ and a $C$ capacitor in series.
Capacitance of one branch $= \dfrac{4C \times C}{4C + C} = \dfrac{4C^2}{5C} = 0.8C$.
Since there are two such branches in parallel, $C_{new} = 0.8C + 0.8C = 1.6C$.

The required ratio is $\dfrac{C_{new}}{C_0} = \dfrac{1.6C}{C} = 1.6$.

Answer: $1.6$