The question is asking for the product and the answer is C. My expected answer was E. Here's how I tried to solve it: Since Br (the leaving group) is at the equatorial position, the structure would 'ring flip'.
Since $\ce{CH3}$ is pointing downwards, I assumed it would have methyl group point outwards (dashed), however, the correct answer is a wedge.
After elimination, how would we know whether the substituent would be pointing inwards, outwards, or even right on the plane? (no wedge or dash)
You diaxial intermediate is good and you can clearly see that there is no way to achieve a Zaytsev product in an $\mathrm{E2}$ elimination since that would require eliminating the methyl group. Carry on with the elimination and then you have a double bond in the top half of the molecule and the methyl group pointing downards — molecule E if inverted by a vertical plane of symmetry traversing the $\ce{C-Me}$ bond.
But the product molecules are drawn with the double bond on the bottom. Thus, you need to rotate the molecule by $180^\circ$. If you do this with a rotation axis in the paper plane (again one parallel to the $\ce{C-Me}$ bond), the double bond is in the correct position but the methyl group is pointing upwards.
You can verify this by determining the absolute configuration of the ipso-methyl carbon: it should be (R) according to your chair drawing and it is (R) in answer C.
