The width is probably meant into the third dimension, like thickness? This means that we can just look at a 2D problem with a surface mass density $\rho$.
The question does not state whether there is friction involved. If there was no friction, every body would slide down exactly the same. This way the question is not interesting in any way.
Since all the “wheels” are supposed to have the surface area and width, they all have the same mass. So one can reword the question perhaps like this: If we redistribute the mass from a circular shape to something else, what changes? The inertia tensor will change. We are only interested in the one moment of inertia which is along the rotation axis. This is given by the following:
$$ I = \iint \mathrm d^2 x \, \rho (\vec x) \, |\vec x^2| = \int \mathrm d x_1 \int \mathrm dx_2 \,\rho (\vec x) \, (x_1^2 + x_2^2) =\int \mathrm d r \, r\int \mathrm d \phi \,\rho (r, \phi) \, r^2 \,.$$
We assume $\rho$ to be constant within the surface and zero outside of it. And there you can see why the circle is the best option: If you have a compact circle, the factor $r^3$ will only become as large as the radius of the circle. If you distort the same surface area to an ellipse, for some angles $\phi$ you will not have that large of $r^3$, that “saves” you a bit of inertia. However, for other angles you need to integrate over $r$ further, and that $r^3$ bit becomes “expensive”.
So no matter how you deviate from the circle with the constraint that the total surface area is the same, you will have make $I$ larger than before. Therefore the circle has the smallest moment of inertia and can rotate the fastest given the same driving moment.
An ellipse with the same surface area has a larger circumference. Therefore it does not need to rotate as fast as the circle to cover the same distance. However, I think that the effect of $I$ becoming larger will outweigh the reduction in revolutions needed to reach the bottom of the slope.