The major product of the following reaction is: 

The reaction shown is an oxymercuration-demercuration reaction.

This reaction converts alkenes into alcohols according to:

$$\mathrm{Markovnikov's\ Rule}$$

The reaction proceeds through a cyclic mercurinium ion intermediate instead of a free carbocation.

Hence, carbocation rearrangements such as hydride or methyl shifts do not occur.

In the first step, $$\mathrm{Hg(OAc)_2/H_2O}$$ adds across the double bond and water attacks the more substituted carbon.

In the second step, $$\mathrm{NaBH_4}$$ replaces the mercury group with hydrogen.

For $$\mathrm{3,3\text{-}Dimethyl\text{-}1\text{-}butene}$$, the hydroxyl group $$\mathrm{(-OH)}$$ attaches to the secondary carbon, while hydrogen attaches to the terminal carbon.

The adjacent $$\mathrm{tert\text{-}butyl}$$ group remains unchanged because no rearrangement occurs.

The final major product is:

$$\mathrm{3,3\text{-}Dimethyl\text{-}2\text{-}butanol}$$

Its structural formula is:

$$\mathrm{(CH_3)_3C-CH(OH)-CH_3}$$. Thus, the right option is A.