Answer
(a) $f_0 = 200~kHz$
(b) $f_0 = 141~kHz$
Work Step by Step
We can write an expression for the resonance frequency:
$\omega_0 = \frac{1}{\sqrt{L~C}}$
$2\pi~f_0 = \frac{1}{\sqrt{L~C}}$
$f_0 = \frac{1}{2\pi~\sqrt{L~C}}$
(a) Note that the resonance frequency $f_0$ does not depend on the resistor value.
$f_0 = 200~kHz$
(b) If the capacitor value is doubled, then the resonance frequency $f_0$ decreases by a factor of $\frac{1}{\sqrt{2}}$
$f_0 = (\frac{1}{\sqrt{2}})~(200~kHz) = 141~kHz$
