Barbara needs two question, no matter what $k$ is. It is clear that one question is never enough since she just learns the sum of some linear combination of $a_1,\dots,a_n$ containing at least two of the variables and such an equation has always no or infinitely many integer solutions for each of the variables. So we need at least two questions. Clearly, if $k \ge 2$ we can just choose two paths which only differ by two cells with number $a_k$ and so we can determine $a_k$ with two questions. The harder part is when $k=1$. Let us first move to the right, then almost completely down until $a_{n-1}$, then right and up again until $a_3$, then right and down to $a_{n-1}$, then right and up to $a_4$, right to $a_5$, down to $a_{n-1}$, right and up to $a_6$ etc. and only in the last three columns go down completely to $a_n$. This way we get $a_1$ plus $3a_n$ plus the sum of the $(n-1) \times (n-1)$-rectangle on the upper right i.e. we learn about \[a_1+2a_2+4a_3+6a_4+\dots+4038a_{2020}+4040a_{2021}+2024a_{2022}.\]On the other hand, we can go right, then twice down, one to the right, three down, one right, two up, one right, three down, one left, one down, two right, three up, one right, four down, two left, one down, three right, four up, one right etc. This way we get \[a_1+a_2+2a_3+3a_4+\dots+2019a_{2020}+2020a_{2021}+1012a_{2022}\]and it is clear that from here we can compute $a_1$.