Answer
$\epsilon = 0.75 = 75 \%$
Work Step by Step
The heat absorbed during the process $A \rightarrow B$ is given by
$Q_H = nC_p \Delta T $
$Q_H = n(\frac{5}{2}) T_A(\frac{T_B}{T_A} - 1) $
$Q_H = nRT_A (\frac{5}{2})(2-1) $
$Q_H = p_oV_o \frac{5}{2}$
The heat released in the process $C \rightarrow D$ is given by
$Q_H = nC_p \Delta T $
$Q_H = n(\frac{5}{2}) T_D(1 - \frac{T_L}{T_D}) $
$Q_H = nRT_D (\frac{5}{2})(1-2) $
Note that $T_D = \frac{1}{4}T_A$
$Q_H = -\frac{1}{4}p_oV_o (\frac{5}{2})$
So
$\epsilon = 1 - |\frac{Q_L}{Q_H}|$
$\epsilon = 1 - |\frac{1}{4}|$
$\epsilon = 0.75 = 75 \%$
