If you got an absorbance of 0.3605, the concentration of the liquid that you measured is 0.0355 ppm.

Depending on the pretreatment you did (taken sample mass, dilutions...) you have to recalculate the obtained concentration to the real one in mg/g.

I do not do dilution.

I do following method from hagendoorn (1994), like below

anthraquinone determination

" For AQ determination 0.02-0.2 g FW was boiled in 5 ml 80% aqueous ethanol for 10 min. After centrifugation (1500 g, 5 min) the supernatant was collected and the pellet was again boiled in 5 ml 80% aqueous ethanol. The supernatants were pooled and made up to 10 ml. The absorption at 434 nm "

You extracted the anthraquinone in 10 ml of ethanol. As the concentration was 0.0355 ppm (mg/l or microg/ml), you must multiply x10 this result, so you have 0.355 microg of anthraquinone. If you divide 0.355 between the mass you weighed (in grams) you will have your final concentration result in microg/g unit

so the result is 0,01775 mg/g because the mass is 0,02 g. very small

Yes, that's the result!!