Since the electric field is a conservative field, we know that the work done by the field is equal to the negative change in potential energy
or, per unit charge
Since the charge starts at $X$ and ends at $Y$, $\Delta V=V_y-V_x$. Therefore
The answer does not depend on how $V_x$ compares to $V_y$. We are told the charge moves from $X$ to $Y$, and that is all we need to know to determine the work done by the field. The given explanation is a nice consistency check, assuming the charge moves only due to the field alone, but it is not needed to actually answer the question. It is a very poor explanation, in my opinion, as it only argues why the work should be positive, but it does not necessarily explain why it should be the exact value of $V_y-V_x$.
I think you are getting confused with other explanations of work done by other forces; these are explanations I really do not like when I see them in introductory materials, as new students typically don't pick up the finer details and just get confused.
When a conservative field does work, we can either just talk about the work it does directly, or we can express that work in terms of a change in potential energy (as done above). When a conservative field does positive work the change in potential energy is negative and vice versa.
However, typically you will see explanations of "moving charges" and talk about the "work done on the charge by an external force". We additionally assume the charge is moved very slowly so that the external force is equal and opposite to the electric force at all instants in time. Then the work done by the external force is equal to the change in potential energy of the charge. And at this point you would be correct that
In this case the problem just asks about the electric field and makes no reference to an external force though. All you need is the start and end potentials to determine the work done by the field.