Work done by a Carnot engine operating between temperatures 127 ° C and 27 ° C is 2   kJ .…

The temperatures in Kelvin scales are 

$$\begin{gathered} T_2=27 °\mathrm{C}+273=300 \mathrm{K} \\ {T}_1=127 °\mathrm{C}+273=400 \mathrm{K} \end{gathered}$$

The efficiency of the Carnot engine is

 $\eta =1-\frac{T_2}{T_1}=1-\frac{300}{400}=\frac{1}{4}$
Using the relation, $\eta =\frac{W}{Q}$
$\Rightarrow \frac{2 \mathrm{kJ}}{Q}=\frac{1}{4}\Rightarrow Q=8 \mathrm{kJ}$