Answer is: molal concentration of glucose in this solution is 6.953 mol/kg.
Kb(H₂O) = 0.512 °C/m.
T(glucose) = 103.56 °C.
T(H₂O) = 100°C.
ΔT = 103.56°C - 100°C.
ΔT = 3.56°C.
ΔT = m(glucose) · Kb(H₂O).
m(glucose) = ΔT ÷ Kb(H₂O).
m(glusose) = 3.56°C ÷ 0.512 °C/m.
m(glucose) = 6.953 mol/kg.
m - molal concentration.