The potential difference between two points, A & B, in an E-field is the negative of the work per unit charge to move a charge from A to B.

V(B) - V(A) = - (1/q)Work Done by E (A to B)

You can take any path from A to B . The easiest path is from A straight down 3 cm (the work is zero on this path because the displacement is perpendicular to force qE). Then straight across 7cm to B. (the work is qEX = qE(.07) for this part since the force is parallel to the displacement).

V(B) - V(A) = - (1/q)(0) - (1/q)qE(.07)

V(B) - V(A) = - (.07)E

Since the difference is negative it means V(A) > V(B)