I thought $B$ because
- All the methyls are equatorial and equatorial groups are more stable.
- All the methyl groups look as spread out as much as possible, reducing steric repulsion.
But the answer is $A$.
Although $A$,$B$ and $D$ could all be said to represent the chair form, the best depiction will show the methlys repelling one another as far as possible. The arrangement in $A$ with two equatorial and one axial methyls makes A the best depiction (textbook).
Also, when converting Haworth to Chair, do we have to match the upward facing groups on the Haworth to upward facing positions on the chair?
I.e. the Haworth has $2$ upward methyls and $1$ downward facing methyl. But in $A$, the left equatorial one is facing upward (as axial in that position is downward), the middle axial methyl is upward, and the right equatorial methyl is downward (as axial in that position is upward). This doesn't match the Haworth.