You are on the right track, but you can solve this one much quicker by using the dilution factor.
The dilution factor is essentially a measure of how concentrated the initial solution was compared with the diluted solution.
Now, the important thing to keep in mind about dilutions is that the amount of solute must remain constant. This implies that the decrease in concentration must be matched by an increase in volume.
This is where the dilution factor comes into play
#color(blue)(ul(color(black)("DF" = c_"concentrated"/c_"diluted" = V_"diluted"/V_"concentrated")))#
In your case, the concentration of the solution must decrease by a factor of
#"DF" = (20color(red)(cancel(color(black)(%))))/(15color(red)(cancel(color(black)(%)))) = color(blue)(4/3)#
This implies that the volume of the solution must increase by factor of
#V_"diluted" = "DF" * V_"stock" #
In your case, you will have
#V_"diluted" = color(blue)(4/3) * "1.5 kg" = "2.0 kg"#
Therefore, you can say that adding
#color(darkgreen)(ul(color(black)("mass of water" = "2.0 kg" - "1.5 kg" = "0.50 kg")))#
of water to
I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the concentration of the diluted solution.
Let's double-check this result by suing your approach. So, you know that the starting solution contains
Since the diluted solution must contain the same amount of sulfuric acid, you can say that
In other words, you need to have
#300 color(red)(cancel(color(black)("g H"_2"SO"_4))) * overbrace("100 g solution"/(15color(red)(cancel(color(black)("g H"_2"SO"_4)))))^(color(blue)("= 15% m/m solution")) = "2000 g solution"#
As you can see, you will once again need to add