A parallel plate capacitor has a dielectric slab of dielectric...

The given arrangement can be considered as two capacitors connected in parallel.

Let the total area of the plates be $A$ and the distance between them be $d$.

The capacitance of the portion with the dielectric (covering $1/3$ of the area) is:
$C_1 = \dfrac{K \epsilon_0 (A/3)}{d} = \dfrac{K \epsilon_0 A}{3d}$

The capacitance of the remaining portion (covering $2/3$ of the area) is:
$C_2 = \dfrac{\epsilon_0 (2A/3)}{d} = \dfrac{2 \epsilon_0 A}{3d}$

The total capacitance $C$ is the sum of $C_1$ and $C_2$ since they are in parallel:
$C = C_1 + C_2 = \dfrac{K \epsilon_0 A}{3d} + \dfrac{2 \epsilon_0 A}{3d} = \dfrac{\epsilon_0 A}{3d}(K + 2)$

Now, finding the ratio $\dfrac{C}{C_1}$:
$\dfrac{C}{C_1} = \dfrac{\dfrac{\epsilon_0 A}{3d}(K + 2)}{\dfrac{K \epsilon_0 A}{3d}} = \dfrac{K + 2}{K}$

Checking other parameters:
Since the two portions are in parallel, the potential difference $V$ across them is the same. The electric field in both regions is $E = \dfrac{V}{d}$. Therefore, $E_1 = E_2$, which gives $\dfrac{E_1}{E_2} = 1$.

The charges on the two portions are $Q_1 = C_1 V$ and $Q_2 = C_2 V$.
The ratio of charges is $\dfrac{Q_1}{Q_2} = \dfrac{C_1}{C_2} = \dfrac{\dfrac{K \epsilon_0 A}{3d}}{\dfrac{2 \epsilon_0 A}{3d}} = \dfrac{K}{2}$.

Answer: $\dfrac{C}{C_1}=\dfrac{2+K}{K}$