Answer
$f_0' = 200~kHz$
Work Step by Step
We can write an expression for the resonance frequency:
$\omega_0 = \frac{1}{\sqrt{L~C}}$
$2\pi~f_0 = \frac{1}{\sqrt{L~C}}$
$f_0 = \frac{1}{2\pi~\sqrt{L~C}}$
We can find the resonance frequency if the capacitor value is doubled, and the inductor value is halved:
$f_0 '= \frac{1}{2\pi~\sqrt{(0.5~L)~(2C)}}$
$f_0' = \frac{1}{2\pi~\sqrt{L~C}}$
$f_0' = 200~kHz$
