Chapter 20 - Electric Potential and Electric Potential Energy - Problems and Conceptual Exercises - Page 720: 53

$0.18\mu F$

We know that $A=Lw$ $\implies A=(5.4m)(6.3\times 10^{-2}m)=34.02\times 10^{-2}m^2$ Now, we can find the required capacitance as follows: $C=\frac{K\epsilon_{\circ}A}{d}$ We plug in the known values to obtain: $C=\frac{(2.1)(8.85\times 10^{-12}F/m)(34.02\times 10^{-2}m^2)}{0.035\times 10^{-3}m}$ $\implies C=0.18\times 10^{-6}F=0.18\mu F$