**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%#