Product formed is:

The correct answer is option 3.

The name of the initial reactant is 3-Methyl-3-phenylbut-1-ene. The reaction of 3-methyl-3-phenyl but-1-ene with \(H_3O^+\) (acidic water) is an example of hydration following the Markovnikov rule. The reaction will be as follow:

Mechanism:

The double bond in 3-methyl-3-phenyl but-1-ene is protonated by the hydronium ion (\(H_3O^+\)). According to Markovnikov's rule, the proton will add to the carbon with the most hydrogen atoms, which in this case is the terminal carbon of the double bond (carbon 1). This leads to the formation of a less stable secondary carbocation. To make the intermediate more stable the methyl group from the third carbon is shifted to the second carbon making a tertiary carbocation which is more stable.
Water (as a nucleophile) attacks the carbocation at carbon 3, leading to the formation of an oxonium ion (protonated alcohol). Finally, deprotonation of the oxonium ion occurs, resulting in the formation of the alcohol.