How many grams of glucose must be added to 275 g of water in order to prepare percent by mass concentration of aqueous glucose solution 1.30% ?

The idea here is that the solution's percent concentration by mass tells you the number of grams of solute, which in your case is glucose, present for every #"100. g"# of the solution.

A solution that is #"1.30%# glucose by mass will contain #"1.30 g"# of glucose for every #"100. g"# of the solution.

So the question is actually asking about the number of grams of glucose that must be added to #"275 g"# of water in order to get #"1.30 g"# of glucose for every #"100. g"# of this solution.

If you take #x# #"g"# to be the mass of glucose needed to make this solution, you can say that after you add the glucose to the water, the total mass of the solution will be

#x quad "g" + "275 g" = (x + 275) quad "g"#

So you know that #x# #"g"# of glucose in #(x + 275)# #"g"# of the solution must be equivalent to #"1.30 g"# of glucose in #"100. g"# of the solution.

This means that you have

#(x color(red)(cancel(color(black)("g glucose"))))/((x + 275) color(red)(cancel(color(black)("g solution")))) = (1.30 color(red)(cancel(color(black)("g glucose"))))/(100. color(red)(cancel(color(black)("g solution"))))#

Rearrange and solve for #x# to get

#100. * x = 1.30 * x + 1.30 * 275#

#98.7 * x = 357.5 implies x = 357.5/98.7 = 3.6221#

Rounded to three sig figs, the number of sig figs you have for your values, the answer will be

#color(darkgreen)(ul(color(black)("mass of glucose = 3.62 g")))#