We can solve this problem using some molarity calculations:
#"molarity" = "mol solute"/"L soln"#
We should convert the given mass of #"(NH"_4")"_2"SO"_4# to moles using its molar mass (calculated to be #132.14# #"g/mol"#):
#10.8cancel("g (NH"_4")"_2"SO"_4)((1color(white)(l)"mol (NH"_4")"_2"SO"_4)/(132.14cancel("g (NH"_4")"_2"SO"_4))) = color(red)(0.0817# #color(red)("mol (NH"_4")"_2"SO"_4#
This is the quantity present in #100# #"mL soln"#, so let's calculate the molarity of the solution (converting volume to liters):
#"molarity" = (color(red)(0.0817)color(white)(l)color(red)("mol (NH"_4")"_2"SO"_4))/(0.100color(white)(l)"L soln") = color(green)(0.817M#
#10# #"mL"# of this solution is added to #50# #"mL H"_2"O"#, which makes a #60##"mL"# total solution.
We can now use the dilution equation
#M_1V_1 = M_2V_2#
to find the molality of the new, #60##"mL"# solution:
#(color(green)(0.0817M))(10color(white)(l)"mL") = (M_2)(60color(white)(l)"mL")#
#M_2 = ((color(green)(0.817M))(10cancel("mL")))/(60cancel("mL")) = 0.136M#
This means that there are #0.136# moles of #"(NH"_4")"_2"SO"_4# per liter of solution.
Let's recognize that #1# #"mol (NH"_4")"_2"SO"_4# contains

#2# #"mol NH"_4^+#

#1# #"mol SO"_4^(2)#
The concentrations of each ion is thus
#(2)(0.136M) = color(blue)(0.272M# #color(blue)("NH"_4^+#
#(1)(0.136M) = color(purple)(0.136M# #color(purple)("SO"_4^(2)#