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"
