We can use the Ideal Gas Law to solve this problem.
#color(blue)(|bar(ul(color(white)(a/a)PV = nRT color(white)(a/a)|)))" "#
Since #"moles" = "mass"/"molar mass"# or #n = m/M#, we can write
#PV = m/MRT#
We can rearrange this to
#PM = m/VRT#
But #"density"= "mass"/"volume"# or #color(brown)(|bar(ul(color(white)(a/a)ρ = m/Vcolor(white)(a/a)|)))" "#
∴ #PM = ρRT# and
#color(blue)(|bar(ul(color(white)(a/a)ρ = (PM)/(RT)color(white)(a/a)|)))" "#
STP = 1 bar and 0 °C, and the molar mass of #"SF"_6# is 146.06 g/mol.
∴ #ρ = (1 color(red)(cancel(color(black)("bar"))) × "146.06 g"·color(red)(cancel(color(black)("mol"^"-1"))))/("0.083 14" color(red)(cancel(color(black)("bar")))·"L"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K")))) = "6.43 g/L"#