Bromo-migration

$\ce{LDA}$ first strips the bromine away to form a double bond; which happens to be a terminal double bond due to $\ce{LDA}$ being a bulky reagent $$ $\ce{Diazomethane}$ forms a $\ce{cyclopropane}$ group between the double bonded carbons and the carbon on $\ce{diazomethane}$ . $$ Finally, the substituted $\ce{cyclopropane}$ is opened by stripping a hydrogen from $\ce{HBr}$ ; note, cyclopropane does this because it exhibits alkene qualities. A $2^{\circ}$ carbocation forms and rearrangment occurs to form a more favorable $3^{\circ}$ carbocation. The $\ce{Br}^-$ comes in to attack the carbocation. $\ce{Answer} = \boxed{A}$