I assume you need to determine the #"w/v"# percent concentration (or m/v) and the #"w/w"# percent concentration (or m/m); since I'm not really sure what m% means, I'll solve for molality as well, just to be on the safe side.
So, in order to get the #"w/v"# percent concentration we need the mass of the solute and the total volume of the solution. We know that the solution has a molarity of #"0.83 mol/dm"^3#; this will allow you to determine how many moles of #Na_2SO_4# are in 1.0 L of solution
#C = n/V => n_(Na_2SO_4) = C * V#
#n_(Na_2SO_4) = 0.83# #"moles"/("dm"^3) * "1 dm"^3 = 0.83# #"moles"#
You can now determine the mass of #Na_2SO_4# by multiplying the number of moles by #"142.0 g/mol"#, the compound's molar mass
#m_(Na_2SO_4) = "0.83 moles" * "142.0"# #"g"/"mol" = 117.9# #"g"#
So, your #"w/v"# percent concentration wil be
#"w/v%" = ("117.9 g")/("1.0 L") * 100% = 11.8%#
Now, if you are looking for the #"w/w"# percent concentration, you have to determine what the mass of the solution is; this is where density comes into play. For the same 1.0 L sample,
#"1.1" "g"/("dm"^3) * "1000 dm"^3 = 1100# #"g solution"#
This means that
#"m/m%" = ("117.9 g")/("1100 g") * 100% = 10.7%#
The solution's molality requires the mass of the solvent in kg, which in this case I assume is water. SInce the total mass of the solution is 1100 g, and the solute weighs 117.9 g, the mass of water is
#m_("water") = "1100 g" - "117.9 g" = "982.1 g"#, therefore
#"m" = ("0.83 moles Na"_2"SO"_4)/("982.1" * "10"^(-3) "kg") = 0.85# #"molal"#
