Explanation

(a) The efficiency$\ \ \eta$ of Carnot engine is defined as the amount of work divided by the heat transferred between the system and the hot reservoir.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \eta = \frac{\Delta W}{\Delta Q_{H}} = 1 - \frac{Tc\ }{T_{H}}$
Where, $T_{C}\ and\ T_{H}$ are temperatures of cold and hot junctions respectively.
Ist case $T_{2} = 0℃ = 0 + 273 = 273K$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_{1} = 200℃ = 200 + 273 = 473\ K$
$\therefore\ \ \ \eta_{1} = 1 - \frac{273}{473} = \frac{200}{473} = 0.4228 \approx 0.423$ …(i)
IInd case
$T_{2} = - 200℃ = - 200 + 273 = 73K$
$T_{1} = 0℃ = 0 + 273 = 273\ K$
$\eta_{2} = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{73}{273} = \frac{200}{273} = 0.732$ …(ii)
From Eqs. (i) and (ii), we get
$\frac{\eta_{1}}{\eta_{2}} = \frac{0.423}{0.732} \approx 0.577$