A parallel plate capacitor with air between the plates has a...

The two capacitors formed are in parallel, hence capacitance of the combination

$C = \dfrac{{C}_{1}{C}_{2}}{{C}_{1} + {C}_{2}}... ( i )$

where, ${C}_{1} = \dfrac{{K}_{1}{ε}_{0}A}{d3} \left(∵ C = \dfrac{{ε}_{0}A}{d}\right)... ( ii )$

${C}_{2} = \dfrac{{K}_{2}{ε}_{0}A}{2d3}... ( iii )$

${C}_{eq} = \dfrac{\dfrac{3{K}_{1}{ε}_{0}A}{d} \times \dfrac{{K}_{2}{ε}_{0}A \cdot 3}{2d}}{\dfrac{3{K}_{1}{ε}_{0}A}{d} + \dfrac{3{K}_{2}{ε}_{0}A}{2d}}$

$= \dfrac{9{K}_{1}{K}_{2}{ε}_{0}^{2}A^{2}}{2d^{2}} \times \dfrac{d}{{ε}_{0}A \times 18}$

It is given

$\dfrac{{ε}_{0}A}{d} = 9pF$

Using given values

${C}_{eq} = 40. 5pF$