Answer
See the detailed answer below.
Work Step by Step
a) We know that the work done by the gas per cycle is the enclosed area under the $P−V$ graph.
Hence,
$$W_{out}={\rm Area}$$
And we can see that the enclosed area under the PV curve is a right triangle.
Thus,
$$W_{out}=\frac{1}{2}(200−100)\times 10^{−6}(400−200)\times 10^3$$
$$W_{out}=\color{red}{\bf 10}\;\rm J$$
We know, for a heat engine, that
$$Q_H=Q_C+W_{out}$$
and we are given $Q_H$ in the given graph.
Thus,
$$Q_C=Q_H-W_{out}=(84+30)-(10)$$
$$Q_C=\color{red}{\bf 104}\;\rm J$$
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b) We know that the thermal efficiency is given by
$$\eta=\dfrac{W_{out}}{Q_H}=\dfrac{10}{(84+30)}$$
$$\eta=\color{red}{\bf 0.088}$$
