Let $A$ be the area of the plates and $d$ be the separation between them. The original capacitance is $C = \frac{\epsilon_0 A}{d}$.

When a dielectric of width $\frac{d}{3}$ is introduced, we can consider the capacitor as two capacitors in series: one with dielectric and one without.

The thickness of the dielectric slab is $\frac{d}{3}$, and its dielectric constant is $k=3$. The capacitance of this part is $C_1 = \frac{k \epsilon_0 A}{d/3} = \frac{3k \epsilon_0 A}{d} = \frac{9 \epsilon_0 A}{d} = 9C$.

The remaining part has a thickness of $d - \frac{d}{3} = \frac{2d}{3}$. The capacitance of this part is $C_2 = \frac{\epsilon_0 A}{2d/3} = \frac{3 \epsilon_0 A}{2d} = \frac{3}{2}C$.

The equivalent capacitance $C'$ of the series combination is given by $\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{9C} + \frac{1}{\frac{3}{2}C} = \frac{1}{9C} + \frac{2}{3C} = \frac{1 + 6}{9C} = \frac{7}{9C}$.

Therefore, $C' = \frac{9}{7}C$.