Respected experts, In this question, in the equation in the box, which axis have they chosen for calculating torque ?

Dear student,
here we have discuused about the marked portion in the question, where you have doubts;
$$\begin{gathered} For pure rolling, \\ \tau = I\alpha \\ \Rightarrow r\times F=I\alpha \\ \Rightarrow r\times T+r\times T= I\alpha \\ \Rightarrow 2rT= I\alpha (as the angle between r and T is {90}^0, r\times T=rT\sin 90=rT) \\ now I = moment of inertia of the cylinder about its axis=\frac{1}{2}M{r}^2 \\ and \alpha = angular acceleration =\frac{a}{r} \\ hence, \\ 2rT=\frac{1}{2}M{r}^2\times \frac{a}{r} \\ \Rightarrow 2T=\frac{1}{2}Ma \\ \Rightarrow T=\frac{1}{4}Ma \end{gathered}$$
So, the cylinder is rotated about the axis, so the  moment of inertia of cylinder about the axis = (1/2)Mr²

Regards,