Right option is (d) 0.55 kN/m^2
For explanation I would say: As per IS 875, for roofs of slope greater than 10^o, the imposed load is reduced by 0.02 kN/m^2 for every degree rise in slope.
Therefore, Imposed load = 0.75 – 0.02*(20^o-10^o) = 0.55 kN/m^2.