Answer
Increased.
Work Step by Step
We know that
$f=\frac{1}{2\pi \sqrt{LC}}$
We plug in the known values to obtain:
$f=\frac{1}{2\pi \sqrt{(0.15mH)(0.20mF)}}$
$f=920Hz$
Since $1.0KHz$ is above the resonance frequency, the frequency should be increased in order to reduce the current.
