The voltage divider created by the Thevenin impedance -j20k and the
input impedance 10k brings 45% of the Thevenin emf to the amplifier input:
10/sqrt(102+202) = 0.45
To get 99%, the Thevenin reactance must be lowered to satisfy the condition:
Rin/sqrt(Rin2+XC2) = 0.99
That results in XC =0.14 Rin = 1.4k.
To get a reactance of 1.4k from a 500 pF capacitor requires a frequency of 227 kHz.