3-ethyl-2-methylpentane-2,3-diol with sulfuric acid
Edit: This answer is based on a typo in the original question referring to a different starting compound than the OP intended.

A quick analysis of this reaction gives me the following:



3-ethyl-2-methylpentan-1,2-diol can be protonated by acid to produce cation A or B. Either of these cations can lose water to form a carbocation


The carbocation formed from A is a tertiary carbocation C that might not be likely to rearrange, but we'll consider the possibility in a bit.
The carbocation formed from B is a primary(!) carbocation D that seems unlikely to form.



  Pinacol rearrangement from cation A




A pinacol rearrangement from cation A through carbocation C forms an aldehyde: 3-ethyl-2-methylpentanal. 


  Pinacol rearrangement from cation B




Having established that carbocation D is unlikely to form, we might suspect no rearrangement, but the rearrangement can occur in a concerted fashion with the loss of water. The end result is a ketone: 4-ethyl-3-hexanone. These kinds of concerted rearrangements to avoid primary cations are also known to occur in Friedel-Crafts reactions.

Neither product seems to match what you provided as the book's answer. Is it possible you miss-named the product?


  Which product is more likely the major product?


Rather than tell you, I will give you a hint:

One rearrangement produces a thermodynamically more stable cation than the other.