You would need #"0.078 g"#, or #"78 mg"# of #"Na"_2"SO"_4#.
The key factor in solving this problem is the balanced chemical equation for the dissociation of #"Na"_2"SO"_4# in water:
#Na_2SO_(4(aq)) -> 2Na_((aq))^(+) + SO_(4(aq))^(2-)#
Notice that 1 mole of #"Na"_2"SO"_4# produces 2 moles of #"Na"^(+)# ions.
What a #"25"# ppm solution of #"Na"^(+)# ions means is that you must have #"25 mg"#, or #"25" * 10^(-3)"g"#, of sodium ions in #"1 L"#, or #"1000 mL"#, of water. You can determine the moles of sodium ions your solution must contain by using
#"25" * 10^(-3)"g" * ("1 mole Na"^(+))/("23.0 g") = "1.09" * 10^(-3)"moles Na"^(+)#
Since the mole ratio between #"Na"_2"SO"_4# and #"Na"^(+)# is #1:2#, the number of moles of the former you'll need is
#1.09 * 10^(_3)"moles Na"^(+) * ("1 mole Na"_2"SO"_4)/("2 moles Na"^(+)) = 0.55 * 10^(-3) "moles Na"_2"SO"_4#
Therefore, the mass of #"Na"_2"SO"_4# needed is
#"0.55" * 10^(-3) "moles Na"_2"SO"_4 * ("142.0 g")/("1 mole") = 78 * 10^(-3)"g" = "78 mg"#
