At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb=0.52, the boiling point of this solution will be:

The number of moles of water in 100 g = 100/18 = 5.555
Let n be the number of moles of solute.
The relative lowering in the vapour pressure is equal to the mole fraction of solute
(P0 - P)/P0 = X
(760 - 732)/760 = n/(n + 5.555)
0.0368 = n/(n + 5.555)
27.144 = n + 5.555
n = 5.555/26.14 = 0.212
The molality of the solution is the number of moles of solute in 1 kg of water
100 g of water corresponds to 0.1 kg
m = 0.212/0.1 = 2.12
The elevation in the boiling point is
ΔTb = kbm = 0.52 × 2.12 = 1.1 °C
The boiling point of the solution is
Tb = 100 + 1.1 = 101.1 °C ≈ 101 °C
Hence, the correct option is A