Intramolecular lactonisation in an unsaturated carboxylic acid
The reaction as you are given turns up no results in SciFinder, meaning it has not been reported in any scientific journal. It is thus the imagination of whoever set the test.

However, closely related reactions are reported and these may come as no surprise to the advanced chemist. They mainly differ in that iodine $\ce{I2}$ is added to the reaction mixture. The mechanism is called iodolactonisation.

Basically, iodine is attacked by the double bond to form an iodonium ion (whether or not this is actually a three-membered ring shall be deemed irrelevant in the context of this answer). This is then rear-side attacked in an intramolecular $\mathrm{S_N2}$ reaction by the deprotonated carboxyl group to form the five-membered lactone kinetically. In a subsequent step, the iodo substituent is removed in an $\mathrm{E2}$ type reaction with a base to regenerate the double bond; due to steric constraints relating to the required anti-periplanar transition state, this newly-formed double bond is shifted along the six-membered ring.



This is described for your compound multiple times in the literature using different bases, for example by Khanjin et al.[1] They use DBU as a milder alternative.

Reference

[1]: N. A. Khanjin, J.P. Snyder, F. M. Menger, J. Am. Chem. Soc. 1999, 121, 11831–11846. DOI: 10.1021/ja992453d.