Step 1: Restate the split rules before touching the tree.
A leaf here can hold at most 2 keys, so a 3rd key forces a leaf split. An internal node can hold at most 2 keys (since it can carry at most 3 child pointers), so a 3rd key at that level forces the node to split too, with its middle key promoted upward instead of being copied.
Step 2: Locate the target leaf for key 3.
The root separators are 5 and 8. Using the ordering already visible from the leaves, the leftmost child [1, 5] is the one whose largest stored key is 5, so any key at or below 5, including 3, lands there.
Step 3: Grow the leaf and split it.
[1, 5] plus the new key 3 gives the ordered list [1, 3, 5]. Splitting an odd list of 3 into two leaves of size 2 and 1 gives [1, 3] on the left and [5] on the right.
The separator carried up to the parent for this new pair is the boundary value 3, the last key of the left half.
Step 4: Count pointers at the root before deciding if it overflows.
The root started with exactly 3 children (its allowed maximum). Inserting the new separator 3 would raise the child count to 4 pointers with keys [3, 5, 8], one pointer past the limit, so the root must be split next.
Step 5: Promote the middle key of the overflowed root.
List the three root keys in order: 3, 5, 8. The middle entry is 5. In an internal-node split this middle entry alone becomes the new root, while 3 stays with the left half and 8 stays with the right half; neither child keeps a copy of 5.
Conclusion:
The tree now has a new single-key root: 5. That is choice (A).\[ \boxed{5} \]