Right option is (a) 10.54 KW

To explain: Thermal efficiency, η = \(\frac{Brake\, power}{Fuel\, consumption*Density}\)*100

BP = \(\frac{FC*CV*\eta}{100} \)

BP = \(\frac{5*10^{-3}*0.836}{3600}*\frac{45.4*10^6*30}{100} \)

BP = 15.814 KW

DP of the tractor = 15.814*0.90*0.75

= 10.67KW.