NOTE: I'll assume the compound is sodium bicarbonate, "NaHCO"_3, not "NaHCD"_3 (even though for this problem it truly doesn't matter).
We're asked to find the percent by mass of "NaHCO"_3 given that the masses of solute and water present are 20 "g" and 600 "g" respectively.
To do this, we can use the equation
%"mass solute" = "mass of solute"/"mass of solution" xx 100%
We known the mass of solute is 20 "g".
The mass of the solution is
overbrace(20color(white)(l)"g")^("mass of NaHCO"_3) + overbrace(600color(white)(l)"g")^"mass of water" = color(red)(620 color(red)("g soln"
Therefore, we have
%"mass NaHCO"_3 = (20color(white)(l)"g NaHCO"_3)/(color(red)(620)color(white)(l)color(red)("g solution")) xx 100% = color(blue)(ul(3.23%
