I'll start by converting degrees Fahrenheit to degrees Celsius.

#"235 F" = "112.78"^@"C"#

#"240 F" = "115.56"^@"C"#

The equation for boiling point elevation is

#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 sucrose 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.

Water's normal boiling point is #"100"^@"C"#, which means that the boiling point elevation for each solution will be

#DeltaT_(b1) = "112.78"^@"C" - "100.0"^@"C" = "12.78"^@"C"#

#DeltaT_(b2) = "115.56"^@"C" - "100.0"^@"C" = "15.56"^@"C"#

Let's start by determining the molalities of the two solutions

#b_1 = (DeltaT_(b1))/(K_b) = ("12.78"^@"C")/("0.512 "^@"C" * "kg/mol") = "25.0 mol/kg"#

#b_2 = (DeltaT_(b2))/(K_b) = ("15.56"^@"C")/("0.512 "^@"C" * "kg/mol") = "30.4 mol/kg"#

The first solution will have #"25.0 moles"# of sucrose for every #"1 kg"# of water. For simplicity, let's assume we have #"1 kg"# of water. This means that the mass of sucrose in solution will be

#"25.0 moles" * ("342.3 g")/("1 mole") = "8557.5 g"#

Percent concentration by mass is defined as the mass of the solute divided by the total mass of the solution and multiplied by 100. Since we have #"1 kg = 1000 g"# of water, the first solution's concentration by mass is

#"%m" = ("8557.5 g")/("8557.5 g + 1000 g") * 100 = 89.5%#

Likewsie, the second solution will have #"30.4 moles"# of sucrose for every #"1 kg"# of water.

#"30.4 moles" * ("342.3 g")/("1 mole") = "10406 g"#

This solution's percent concentration by mass will be

#"%m" = ("10406 g")/("11406 g") * 100 = 91.2%#