Correct Answer  Option 4 : 16.3 kJ/s
CONCEPT:

Carnot engine an ideal reversible engine that operates between two temperatures T1 (Source), and T2 (Sink).

Carnot engine operates through a series of two isothermal and adiabatic processes called the Carnot cycle.
 The steps of the Carnot cycle are
 Isothermal expansion
 Adiabatic expansion
 Isothermal compression
 Adiabatic compression
 The efficiency of the Carnot engine is defined as the ratio of net work done per cycle by the engine to heat absorbed per cycle by the working substance from the source.
 The efficiency is given by
\(\eta = \frac{W}{Q_1} =\frac{Q_1Q_2}{Q_1} = 1\frac{Q_2}{Q_1}\)
Where W = Work, Q1 = Amount of heat absorbed, Q2 = Amount of heat rejected
As \(\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\)
\( \eta =1\frac{{{T}_{2}}}{{{T}_{1}}}\)
Where T1 = temperature of the source and T2 = temperature of the sink.
SOLUTION:
Given  T1 = 200° C = 473 K, T2 = 20° C = 293 K and W = 10 kW = 10 kJ/sec
 The efficiency is given by
\(⇒ \eta =1\frac{{{T}_{2}}}{{{T}_{1}}}\)
\(⇒ \eta =1\frac{{{T}_{2}}}{{{T}_{1}}}=1\frac{{{293}}}{{{473}}}=0.374\)
 The amount of heat absorbed by the Carnot engine can be calculated is
\(⇒ Q_1= \frac{W}{\eta}=\frac{10}{0.374}=26.73\; kW\)
 The amount of heat rejected to the sink will be:
⇒ Q2 = Q1  W
⇒ Q2 = 26.73  10 = 16.73 kW = 16.73 kJ/sec