What would be the product (B) in the following sequence of reaction?
This is a fun one, and a common question in this course.  First, as you say, H and Br are lost by anti elimination, since they helpfully tell you base and E2 are involved.  That means the H and Br are lost from opposite faces of the future double bond, leaving us with a trans intermediate that has a mirror plane (the plane of the molecule)

Now we need to add Br+ to make a bromonium intermediate.  Notice that both carbons end up as (R) or else both end up as (S), so two enantiomers are possible here.

But, whichever carbon is attacked by Br- ends up with an inversion of symmetry, so you now have two carbons of opposite chirality.  The result, as shown in your answer, is a molecule with a mirror plane.  In other words, adding bromine to opposite sides of a trans double bond makes both sides come out as mirror images, once you turn the bromines around to line up.  This compound is a meso compound.  But ... what about the other isomer if the Br+ originally goes to the other side of the double bond?  Well, just turn your Fisher structure upside down. :)