A rolled capacitor is made of two long strips of foil, each 5.0 cm wide, with a thin film between…

Step 1. The area with air. We want $A$, so $A = \dfrac{Cd}{\varepsilon_0} = \dfrac{2.2 \times 10^{-6}\ \text{F} \times 5.0 \times 10^{-5}\ \text{m}}{8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)} = 12.4\ \text{m}^2$.

Step 2. The length with air. $L = \dfrac{A}{w} = \dfrac{12.4\ \text{m}^2}{0.050\ \text{m}} = 249\ \text{m}$.

Step 3. With the film. $C = \dfrac{K\varepsilon_0 A}{d}$, so $A = \dfrac{Cd}{K\varepsilon_0}$ is $3.0$ times smaller, $4.14\ \text{m}^2$. The strip is $\dfrac{249\ \text{m}}{3.0} = 82.9$ m long.