1. Initial capacitor (air-filled)

For a parallel plate capacitor with air:

$C_0=\frac{\varepsilon_0 A}{d}=9\,\text{pF}$

So,

$\frac{\varepsilon_0 A}{d}=9\,\text{pF}$

2. New arrangement

The dielectric slabs are stacked along the separation between the plates, so they behave like two capacitors in series.

• First slab: dielectric constant $k_1=3$, thickness $d_1=\dfrac d3$
• Second slab: dielectric constant $k_2=6$, thickness $d_2=\dfrac{2d}{3}$

For layered dielectrics along thickness, equivalent capacitance is

$C=\frac{\varepsilon_0 A}{\dfrac{d_1}{k_1}+\dfrac{d_2}{k_2}}$

Substitute values:

$C=\frac{\varepsilon_0 A}{\dfrac{d/3}{3}+\dfrac{2d/3}{6}}$

$C=\frac{\varepsilon_0 A}{\dfrac{d}{9}+\dfrac{2d}{18}}$

$C=\frac{\varepsilon_0 A}{\dfrac{d}{9}+\dfrac{d}{9}}$

$C=\frac{\varepsilon_0 A}{\dfrac{2d}{9}}$

$C=\frac{9}{2}\cdot \frac{\varepsilon_0 A}{d}$

Using $\dfrac{\varepsilon_0 A}{d}=9\,\text{pF}$,

$C=\frac{9}{2}\times 9=40.5\,\text{pF}$

3. Option check

• A: $1.8\,\text{pF}$ — incorrect
• B: $45\,\text{pF}$ — incorrect
• C: $40.5\,\text{pF}$ — correct
• D: $20.25\,\text{pF}$ — incorrect

Therefore, the capacitance is:

$\boxed{40.5\,\text{pF}}$