Answer
$Q_{abc} = 14.8 \space kJ$
Work Step by Step
The energy added to during the stroke of abc is given by
$Q_{abc} = nC_v(T_b-T_a) + nC_p(T_c-T_b) $
$Q_{abc} = n(\frac{3}{2}R) T_a (\frac{T_b}{T_a} - 1) + n(\frac{5}{2}R) T_a (\frac{T_c}{T_a} - \frac {T_b}{T_a})$
$Q_{abc} = nRT_a [\frac{3}{2} (\frac{T_b}{T_a} - 1) + \frac{5}{2} (\frac{T_c}{T_a} - \frac {T_b}{T_a})]$
$Q_{abc} = p_oV_o [\frac{3}{2} (2- 1) + \frac{5}{2} (4- 2)]$
$Q_{abc} = \frac{13}{2} p_oV_o $
Since the work done, $W = p_oV_o = 2.27 kJ$, so
$Q_{abc} = \frac{13}{2} 2.27 kJ = 14755 J $ or $14.8 \space kJ$
