Some math:

every 4 seconds, the 1880uf capacitor is charged up to 380v. So total energy consumed by R1 and stored in C is 0.5*C*V^2 = 140j. Or the power dissipation over R1 is 140j/4s=30w (the average figure). The instantenous power can be as high as 3Kw.

So a 6w resistor sounds like undersized here.

 

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every 4 seconds, the 1880uf capacitor is charged up to 380v. So total energy consumed by R1 and stored in C is 0.5*C*V^2 = 140j. Or the power dissipation over R1 is 140j/4s=30w (the average figure). The instantenous power can be as high as 3Kw.
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The simulation showed significantly less power dissipated in R1 than R2.
Either the simulation is wrong or the difference is related to the capacitor being charged by a rectified sine-wave and not a DC level.

This brings up the interesting observation that it's apparently more efficient to charge a capacitor from a rectified AC source than a DC source.
The simulation below shows the energy output of V1 (AC) and V3 (DC) for charging the same size capacitor to the same voltage.
The DC source output energy is nearly 14% higher (which is the added dissipation in the resistor), even with the reduced efficiency of the bridge diodes for the AC source

 

" after a while the capacitors charge and the circuit is in a stable situation a relay bypasses this resistor....... I think in worst situation (when input switch exactly switches in peak of input ac voltage) the stress of R1 may be equal to R2. I put my resistor in R2 position to see if it can withstand in worst input situation."

imijoon;

The stress on R2 overstates the heat stress on R1. It makes no significant or practical difference when R1 is switched into the AC signal because the time constant of the RC circuit is an order of magnitude longer than the period of peak AC voltages.

RC = 88ms
1/4 AC cycle = 5ms

The max voltage on R1 during the first quarter AC cycle is only about 3% less if switched in at 0V versus being switched in at 380V. And that small difference disappears very quickly after the first 5 ms.

Avg of Vp*Sin() pulse = 2/pi*Vp
Percent difference in max voltage seen by R1 in first quarter of AC cycle = (2/pi) * [1- (e^(-5ms/RC)] = 3.5%
note: this is due to charging of capacitor at t= 5ms versus t= 0ms.

It would be even less if you account for the series inductance of the wire wound resistor R1 which will buck the initial current pulse when switched in at peak AC line voltage. So regardless of when it is switched on, R1 will see ~380V peak with no difference in surge power.

A way to test R1 is to test to failure by repeated and more frequent switch-ins of the supply than you expect in practice and do this at maximum resistor temperature - presumably near PCB/ambient temperature in the case. Also change R2 to draw the maximum load this supply will see during ramp up. I doubt you intend to supply 3kW while the voltage ramps up!

I'd get resistors from a quality vender that supplies surge power de-rating curves so you can determine likely candidates from the spec sheet.


edit: My AC period calculation was originally based on 60Hz not 50Hz, so I changed the formula. Also I made some other changes in the text for clarity and to reflect your use of a bypass relay. Looking at several data sheets and resistor app notes from manufacturers I think you will have to find the right size resistor by testing. Pulse loads below 5 seconds are generally not de-rated.

 

So total energy consumed by R1 and stored in C is 0.5*C*V^2 = 140j.

The exact figure from the formula above is 136j. Confirmed by simulation as well.

Or the power dissipation over R1 is 140j/4s=30w (the average figure).

assuming on 4s and off 4s, the total power dissipation over R1 / R2 is 140j/8s = 17w.

Still far greater than 6w.

 

The exact figure from the formula above is 136j. Confirmed by simulation as well.
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Not by my simulation. That gives a value of 101.6J for R1.
What simulation are you referring to?