Answer
$C = 1.3\times 10^{-6}~F$
Work Step by Step
We can find the required capacitor value:
$\omega_0 = \frac{1}{\sqrt{L~C}}$
$2\pi~f_0 = \frac{1}{\sqrt{L~C}}$
$(2\pi~f_0)^2 = \frac{1}{L~C}$
$C = \frac{1}{(2\pi~f_0)^2~L}$
$C = \frac{1}{(2\pi~\cdot 1000~Hz)^2~(0.020~H)}$
$C = 1.3\times 10^{-6}~F$
