Simple LP filter - calculate cutoff frequency
If you regard R2 as being in parallel with R1, you will obtain the correct frequency cut-off. This is just superposition. R1 and R2 produce a value of: -

\$\dfrac{1}{\frac{1}{R_1}+\frac{1}{R_2}}\$ = 1.232k ohms.

This forms a low pass filter with C1 of: -

\$\dfrac{1}{2\pi\times 1.232\times 10^3 \times 560\times 10^{-12}}\$ = 230.686 kHz