A "capacitance multiplier" is definitely a valid way to reduce ripple in a power supply.
It is basically a single-pole R-C low-pass filter followed by an emitter follower as a buffer. Instead of shunting the ripple voltage to ground through a capacitor (creating a ripple current problem), the transistor simply blocks it.
Of course, this does not come without a cost. First of all, you need to make sure that the output voltage is less than the minimum input voltage -- i.e., the lowest voltage that the ripple reaches. This means that you need to add something like a zener diode or a resistor between the transistor base and ground.
This voltage difference needs to be large enough to keep the transistor conducting at all times, which also means that it needs to be large enough to supply the required base current through the resistor in your diagram. Suppose you're willing to live with a minimum voltage drop across the transistor of 2 V.
If the transistor has a current gain of 50, that means you need 60 mA of base current for an output current of 3 A. Since the minimum drop across the resistor is 1.3 V, this means that the resistor can be no larger than 1.3 V / 60 mA = 22 Ω
Any voltage difference between the instantaneous input voltage and the output voltage gets dissipated in the transistor at whatever the load current is. For example, if the ripple is 3 Vp-p on top of a minimum drop of 2 V, the average voltage across the transistor is at least 3.5 V. At a load current of 3A, this becomes a minimum of 10.5 W of wasted power. This represents a significant drop in overall efficiency, which may negate the benefits of using a switching converter in the first place.
There are several confusing aspects to your question, however. If your desired output is 24 V, why do you have 80-100 V at the rectifier? There's a huge mismatch somewhere in your basic design. Also, you talk about a flyback converter, but your diagram shows a transformer connected directly to the mains. Which is it?