Answer
The resonant frequency increases if $C_3$ is removed.
Work Step by Step
We can find the equivalent capacitance of three capacitors in parallel:
$C = C_1+C_2+C_3 = 4.00~\mu F+2.50~\mu F+3.50~\mu F = 10.0~\mu F$
We can find the equivalent inductance of two inductors in series:
$L = L_1+L_2 = 1.70~mH+2.30~mH = 4.00~mH$
We can find an expression for the original resonant frequency:
$\omega_d = \frac{1}{\sqrt{L~C}}$
$2\pi~f_d = \frac{1}{\sqrt{L~C}}$
$f_d = \frac{1}{2\pi~\sqrt{L~C}}$
If $C_3$ is removed, then the equivalent capacitance $C$ decreases, which in turn increases the resonant frequency.
The resonant frequency increases if $C_3$ is removed.
