This question is built on drawn structures, so the options cannot be reproduced as text. The chemistry it tests is set out below.

A carboxylic acid with a carbon-carbon double bond elsewhere in the chain, treated with iodine, undergoes iodolactonisation — a reaction whose product is a cyclic ester (lactone) carrying an iodine atom.

The mechanism runs in two steps. Iodine first adds to the alkene to form a bridged, three-membered iodonium ion. That intermediate is electrophilic at both alkene carbons, and it is opened intramolecularly by the carboxylate oxygen of the same molecule acting as the nucleophile.

Two consequences follow, and they are what the options test.

Regiochemistry. The carboxylate attacks so as to form the more favourable ring, normally the five-membered γ-lactone in preference to a six-membered one, and it opens the iodonium at the carbon better able to bear positive charge — a Markovnikov-type outcome.

Stereochemistry. Because the nucleophile must attack the opposite face from the bridging iodine, ring opening is strictly anti. The oxygen and the iodine therefore end up trans across the newly formed ring, giving a single defined diastereomer.

So the correct option is the iodo-substituted lactone with anti stereochemistry between the C-O and C-I bonds; simple addition products in which the acid remains free, or lactones without the iodine, do not correspond to this mechanism.

Per the official final answer key the answer is option (A).