(b) Efficiency, $\eta = \frac{output\ power}{consuming\ power} \times 100\%$
$Here\ ,\ P_{output} = 10kW$
$P_{input} = 2 \times 10^{3}calg^{- 1} \times gs^{- 1}$
$= 2 \times 10^{3}cals^{- 1}$
$= 2 \times 10^{3} \times 4.2Js^{- 1}$
$8.4kW$
As, $P_{output} > P_{input}$, hence it is never possible.