Before doing any calculation, try to predict what will happen to the boiling poing of the aluminium nitrate solution.
You know that aluminium nitrate,
#"Al"("NO"_3)_text(3(s]) -> "Al"_text((aq])^(3+) + 3"NO"_text(3(aq])^(-)#
This tells you that every mole of aluminium nitrate will produce four moles of ions in solution.
Since boiling-point elevation is a colligative property that depends on how many molecules of solute are present in solution, you can expect the boiling point of the aluminium nitrate's solution to be higher than that of the sugar solution.
Now, with that being said, the formula for boiling-point elevation looks like this
#DeltaT_b = i * K_b * b" "#, where
The van't Hoff factor tells you how many particles you get in solution for every molecule or formula unit of solute that dissolves.
Sugar is a non-electrolyte, which means that its van't Hoff factor is equal to
Therefore, If you add the same number of moles of aluminium nitrate as you add of sugar in the same mass of water, the boiling-point elevation of the aluminium nitrate solution will be four times higher than that of the sugar solution.
#T_"b aluminium nitrate" = 4 xx T_"b sugar"#
#T_"b aluminium nitrate" = 4 * 1""^@"C" = color(green)(4""^@"C")#
Mathematically, you can prove this by using
#K_b * b = (DeltaT_(b1))/i_1 ->#for the sugar solution
For the aluminium nitrate solution, you have the exact same amount of moles of solute and the exact same amount of water, which means that the product
#DeltaT_(b2) = i_2 * (DeltaT_(b_1))/i_1 = i_2/i_1 * DeltaT_(b1)#
#DeltaT_(b2) = 4/1 * 1""^@"C" = 4""^@"C"#