The answer is "213.5 g"213.5 g of glucose.
Start with the equation for boiling point elevation
DeltaT_b = i * K_b * b, where
DeltaT_b - the poiling point elevation;
i - the van't Hoff factor - in your case "i=1" because glucose does not dissociate when dissolved in water;
K_b - the ebullioscopic constant - for water its value is listed as "0.512 "^@"C" * "kg/mol";
b - the molality of the solution.
You know that pure water boils at "100.0"^@"C", which means that the boiling point elevation for this solution is
"102.36"^@"C" - "100.0"^@"C" = "2.36"^@"C"
This means that the solution's molality is
b = (DeltaT_b)/K_b = ("2.36 "^@"C")/("0.512 "^@"C" * "kg/mol") = "4.61 mol/kg"
This means that you have "4.61 moles" of glucose for every "1 kg" of water. The number of moles you'll have in "255 g" of water will be
"255 g" * ("4.61 moles")/("1000 g") = "1.18 moles"
Since you know glucose's molar mass, the mass of glucose will be equal to
"1.18 moles" * ("180.9 g")/("1 mole") = "213.5 g", or
m_("glucose") = "214 g" - rounded to three sig figs.
