Your welcome. For 3 mA(p-p) current you need a total impedance of 20/0.003=6667 ohms. So the total load impedance needs to be 6667-600≈6000 ohms. To get that put a 6.4 uH coil in parallel with a 6000 ohm resistor in parallel with your capacitor. The capacitor and inductor will resonate at 300 KHZ leaving a pure 6000 ohm load impedance, and the 3 mA(p-p) current will flow through the 6000 ohms resistor.
.[if this current will flow through my load too? because 6000 ohm resistor is parallel with load ./QUOTE]