You have to go backwards, from the output voltage and current.
Let's say you want maximum 3.3v at 500mA, for safety go +10%, to 550mA . That's 3.3v x 550mA = 1815 mW
You look in the datasheet and try to determine how efficient that regulator will be converting ~30v to 3.3v at that current amount, with the inductor you choose (again specified in datasheet).
Let's say you determine the efficiency determined is 70% : that means 1815 mW were 70% of the total amount consumed by the regulator, so 100% is 100 * 1815 / 70 = 2592 mW ... let's round it up to 2600 mW.
So if you assume your voltage will be between 24v and 30v... let's go with 24v, then your input current will be 2600 mW / 24 = 108.33 mA
=> the switching regulator needs 24v+ 108mA+ to output 3.3v 500mA with around 70% efficiency
Now, the switching regulator may pulls energy in bursts or whatever, so let's say you design the input for at least 150 mA
This means the input capacitor must be sized so that the voltage will not go lower than your desired voltage.
In your circuit you have a bridge rectifier rectifying 24v AC to DC ... so you'll get peak voltage of Vdc peak = sqrt(2) x Vac - 2 x Vdiode.
For a generic bridge rectifier, you'd have Vdc peak = 1.414 x 24v - 2 x 0.8v = ~ 32v
Because you may have less than 110v / 230v on the mains, lowering the output on the secondary, assume a lower peak DC voltage, let's say your maximum will be 30v DC.
0.8v is a fairly good value for the voltage drop on an internal diode at around 0.1-0.2A ... for around 1-2A, you may want to choose around 1v
So you can use a formula to approximate capacitance required : C (Farads) = Current / [ 2 x AC Frequency x (Vdc peak - Vdc min desired) ]
Let's say you're in US where AC frequency is 60Hz and you want minimum 24v and at least 150mA (0.15) and your peak voltage is 30v, not 32v (for big margins)
=> C = 0.15 A / ( 2 x 60 x (30-24)) = 0.15 / 120 x 6 = 0.00020833 Farads or around 208uF ... this means you could use next value up like 270uF or 330uF and you'll be safe.
You can go with much bigger values but that only means the voltage will be closer to 30-32v for longer periods of time.
Also, bigger capacities are not so good when you start the circuit cold... the empty capacitors act almost like a short sucking in energy, so for a short moment, there's very high current through the transformer, and that can cause the fuse to pop... that's why you often see time delay fuse before the transformer.
Anyway... you have 3.3v up to 550mA , determined about 110mA at 24-30v on input of regulator but used 150mA to determine capacitor, so let's continue to use that.
If you output 24v 150mA on the secondary, that's 3.6 watts (24x0.15).
If your input is 110v, then the primary should take 3.6w / 110 = 0.0327 A or 32 mA ... so in theory, a 50mA fuse would be plenty. But it would probably make more sense to use a 100mA time delay fuse.
You don't need 2 capacitors , if the switching regulator is close to that first capacitor. A single one is plenty.
It may help to have a ceramic 0.1uF right by the input pins of the regulator.
Also you may find cheaper to go with individual diodes for the bridge rectifier... you lose a bit of board space but not much.
Cheapest rectifiers on Digikey are something like 19 cents if you buy 25+, while single diodes are much cheaper.
But you can get much cheaper parts from lcsc.com or tme.eu or other distributors that do less known manufacturers.