The idea here is that you need to use the known composition of the solution to figure out the mass of glucose, the solute, present in exactly
So if you dissolve
#overbrace("7.79 g")^(color(blue)("mass of solute")) + overbrace("237.7 g")^(color(blue)("mass of solvent")) = overbrace("245.5 g")^(color(blue)("mass of solution"))#
So, you know that this solution contains
#100 color(red)(cancel(color(black)("g solution"))) * "7.79 g glucose"/(245.5color(red)(cancel(color(black)("g solution")))) = "3.17 g glucose"#
This means that the solution's percent concentration by mass will be equal to
#color(darkgreen)(ul(color(black)("% m/m = 3.17% glucose")))#
The answer is rounded to three sig figs, the number of sig figs you have for the mass of glucose.