We were given #2.6 xx 10^23# molecules of #"SO"_2#. That means that
#N = 2.6 xx 10^23 \ "molecules"#
Use
#n ("number of moles") = N/(N_A ("Avogadro's constant"))#
to get
#n = (2.6 xx 10^23\ "molecules") / (6.02 xx 10^23\ "molecules/mol") = "0.432 moles"#
Now use
#m = n xx M ("molar mass")#
to get
#m = "0.432 mol" xx "64.066 g/mol" = "27.677 g"#
So
#m ("in kilograms") = "27.677 g" / (10^3 \ "g/kg") = "0.028 kg"#