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.