Possible Solution
Content 
Notes
Actually we get a nice formula due to the fact that indices for rows and columns start at zero:

The \(I\)th row of \(\overline{A}_Z\) belongs to panel \(A_i\) with \(i = \left\lfloor \frac{I}{M_r} \right\rfloor\). Panels \(A_0, \dots, A_{i1}\) have a total of \(i \cdot M_r \cdot K\) elements. These must be stored in the buffer in front of the elements from panel \(A_i\).
So the offset in the buffer is at least \(i \cdot M_r \cdot K = \left\lfloor \frac{I}{M_r} \right\rfloor \cdot M_r \cdot K\)

In the panel \(A_i\) all elements with column index smaller than \(J\) must be stored in front of the elements with column index \(J\). These are in total \(M_r \cdot J\) elements.
So the offset must be increased by \(M_r \cdot J\).

The element \(a_{I, J}\) has the row index \(I \bmod M_r\) in the panel \(A_i\). All elements in the \(J\)th column of \(A_i\) with smaller rowindex must be stored in front of \(a_{I,J}\). These are \(I \bmod M_r\) elements.
So the offset must be increased by \(I \bmod M_r\).
Before you continue
Make sure you understand the formula! :)