On which members the sum operator operates I don't know on which operators the sum operator operates? I was thinking all the operators till the identity matrix (including).
Here is the screenshot of the equation: EQUATION
Here is the link to the paper describing the equation: LINK.
 A: You are right. The scope of the sigma symbol ends with the identity matrix $I_3$.
We have
\begin{align*}
\sum_{i=1}^{n_k}&\left[3-\left({\bf r}_i^T{\bf b}_i\right)^2\right]I_3+\left({\bf b}_i^T{\bf r}_i\right)\left({\bf b}_i^T{\bf r}_i+{\bf r}_i{\bf b}_i^T\right)
+\left({\bf r}_i\times\right)\left({\bf b}_i{\bf b}_i^T\right)\left({\bf r}_i\times\right)^T\\
&=\left(\sum_{i=1}^{n_k}\left[3-\left({\bf r}_i^T{\bf b}_i\right)^2\right]I_3\right)+\left({\bf b}_i^T{\bf r}_i\right)\left({\bf b}_i^T{\bf r}_i+{\bf r}_i{\bf b}_i^T\right)
+\left({\bf r}_i\times\right)\left({\bf b}_i{\bf b}_i^T\right)\left({\bf r}_i\times\right)^T\tag{1}\\
&=\sum_{q=1}^{n_k}\left[3-\left({\bf r}_q^T{\bf b}_q\right)^2\right]I_3+\left({\bf b}_i^T{\bf r}_i\right)\left({\bf b}_i^T{\bf r}_i+{\bf r}_i{\bf b}_i^T\right)+\left({\bf r}_i\times\right)\left({\bf b}_i{\bf b}_i^T\right)\left({\bf r}_i\times\right)^T\\
\end{align*}
The scope of the sum is given by the surrounding parenthesis in (1). The index $i$ of the sigma symbol is a bound variable, whereas the other instances of $i$ are free variables.
But since the terms with the free variable $i$ do not occur elsewhere in the text, we can assume there is simply a typo and parenthesis are missing. We then have
\begin{align*}
\sum_{i=1}^{n_k}&\left(\left[3-\left({\bf r}_i^T{\bf b}_i\right)^2\right]I_3+\left({\bf b}_i^T{\bf r}_i\right)\left({\bf b}_i^T{\bf r}_i+{\bf r}_i{\bf b}_i^T\right)
+\left({\bf r}_i\times\right)\left({\bf b}_i{\bf b}_i^T\right)\left({\bf r}_i\times\right)^T\right)\\
\end{align*}
