From given cyclic process,

Process equation is

$(p-4)^2+(V-4)^2=4 \quad \dots(i)$

Now, from $p V=n R T$

We can say that $T$ is maximum when $p V$ is maximum.

Now, for given cyclic process, $p V$ maximum occur when $p^2 V^2$ is maximum.

Now,

$p^2 V^2=p^2\left(4-(p-4)^2\right)$ [from Eq. $(i)$]

Now, $p^2 V^2$ is maximum when $\frac{d}{d p} p^2 V^2=0$

$\Rightarrow \quad \frac{d}{d p}\left(p^2 \cdot\left(p^2 \cdot\left(4-(p-4)^2\right)=0\right.\right.$

$\Rightarrow \quad p^2-8 p+14=0$

$=4 \pm \sqrt{2}$

and from Eq. $(i)$, we get

$p=4 \pm \sqrt{2}, V=4 \pm \sqrt{2}$

Taking positive values, we have

$(p V)_{\max } \Rightarrow p=4+\sqrt{2}$

and $V=4+\sqrt{2}$

So, by gas equation, we have

$T_{\max } =\frac{(p V)_{\max }}{R} \quad \text { [for } 1 \,mol \text { of gas] }$

$=\frac{(4+\sqrt{2})(4+\sqrt{2})}{R}$

$=\frac{16+2+2 \times 4 \times \sqrt{2}}{R}$

$=\frac{29.32}{R}=\frac{30}{R}$