How to determine the corner frequency needed?
Attenuating to 10% means the dB reduction is 20 dB because 20 log 10 = 20 dB. 

A 2nd order low-pass filter will normally reduce the high frequencies at 40 dB per decade therefore, you need to have the filter's cut-off frequency half a decade below 100 Hz at 31.62 Hz (\$\sqrt{10\times 100}\$). 

Here's an approximation: -



This was modelled with an RLC low pass filter interactive tool having a butterworth response (\$\zeta\$ = 0.7071). Make sure that if you are using a sallen key filter the op-amp has a suitable GBWP or it might not meet your expectations.

Note that I didn't go to great lengths to choose LC to make exactly a cut-off 31.62 Hz but it's near enough to demonstrate that the marker at 100.66 Hz is 20 dB down.