A rectangular dielectric slab has dimensions of 20 cm × 20 cm × 1.0...

The capacitance of a parallel plate capacitor with a dielectric is given by $C = \dfrac{K \varepsilon_0 A}{d}$

Given:
Dielectric constant, $K = 4.0$
Area of the square faces, $A = 20 \text{ cm} \times 20 \text{ cm} = 400 \text{ cm}^2 = 4 \times 10^{-2} \text{ m}^2$
Distance between the plates (thickness), $d = 1.0 \text{ mm} = 10^{-3} \text{ m}$
Permittivity of free space, $\varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m}$

Substituting the values into the formula:
$C = \dfrac{4.0 \times 8.85 \times 10^{-12} \times 4 \times 10^{-2}}{10^{-3}}$

$C = 141.6 \times 10^{-11} \text{ F}$

$C = 1.416 \times 10^{-9} \text{ F} \approx 1.42 \text{ nF}$