Answer
$300\;\rm kPa$
Work Step by Step
We know that the work done by the gas per cycle is the enclosed area under the $pāV$ curve.
Hence,
$$W_{out}={\rm Area}$$
And we can see that the enclosed area under the $P-V$ curve is a triangle.
Hence,
$$W_{out}=\frac{1}{2} hb$$
where $h$ is the triangle height and $b$ is its base.
Plugging from the given graph;
$$W_{out}=\frac{1}{2}(P_{max}-100)\times 10^3 \times (800-200)\times 10^{-6}$$
Hence,
$$P_{max}=\dfrac{2W_{out}}{10^3 \times (800-200)\times 10^{-6}}+100$$
Plugging the known;
$$P_{max}=\dfrac{2 (60)}{10^3 \times (800-200)\times 10^{-6}}+100$$
$$P_{max}=\color{red}{\bf 300}\;\rm kPa$$
