The answer is #"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.
