What volume of the #"0.8-mol L"^(-1)# aluminium nitrate, #"Al"("NO"_3)_3#, solution contains #"85 g"# of the solute?

You know that your solution has a molarity of #"0.8 mol L"^(-1)#, which implies that every #"1 L"# of solution contains #0.8# moles of dissolved aluminium nitrate.

You can actually convert the concentration of the solution to grams per liter by using the molar mass of the solute.

#0.8 color(white)(.)color(red)(cancel(color(black)("moles Al"("NO"_3)_3)))/"1 L solution" * "213.0 g"/(1color(red)(cancel(color(black)("mole Al"("NO"_3)_3)))) = "170.4 g L"^(-1)#

This means that every #"1 L"# of solution contains #"170.4 g"# of dissolved aliuminium nitrate.

Since solutions are homogeneous mixtures, i.e. they have the same composition throughout, you can use the known composition of the solution to calculate the volume that would contain #"85 g"# of dissolved solute

#85 color(red)(cancel(color(black)("g Al"("NO"_3)_3))) * "1 L solution"/(170.4color(red)(cancel(color(black)("g Al"("NO"_3)_3)))) = color(darkgreen)(ul(color(black)("0.5 L solution")))#

The answer must be rounded to one significant figure, the number of sig figs you have for the molarity of the solution.

As a side note, because aluminium nitrate is soluble in water, you can't say that the solution contains #"0.8 mol L"^(-1)# of aluminium nitrate.

Instead, you can say that the solution contains #"0.8 mol L"^(-1)# of aluminium cations, #"Al"^(3+)#, and #3 xx "0.8 mol L"^(-1)# of nitrate anions, #"NO"_3^(-)#.