Cutoff frequency of a second order low pass RLC circuit
Is there another formula?


There is the correct formula (rather than your incorrect formula): -

$$F_C = \dfrac{1}{2\pi\sqrt{LC}}$$

This is the natural resonant frequency for a 2nd order low pass filter and is not necessarily the 3dB cut-off frequency because the 2nd order type filter has the ability to produce a peak in the frequency response. Here's a picture based on your component values: -



Picture from this interactive filter website and notice that at the natural resonant frequency (10.7 kHz) the attenuation is 3.979 dB. If you use the cursor you can find the 3 dB point to be about 8.92 kHz.