How much work is required to move 1 micro coulombs of electric charge a distance of 2 meters along an equipotential line or conductor of 12.5 volts?

1 volt = 1 joule per coulomb

That's the easy part. Now, before we go any farther, we need to
straighten out the language in the question:

An "equipotential line" is a line with every point at the same potential,
so there's no voltage difference along such a line, and it takes no work
for charge to move on it.

Similarly, for discussion purposes, a conductor is assumed to be
lossless, with all points on it at the same potential.  So it takes no
work to move charge along a conductor.

If you really mean an equipotential line or a conductor, then the
answer is:
Zero, no matter how much charge there is, and no matter how far it
moves, and all the numbers, including the '12.5 volts' at the end, are
totally red herrings, just there to confuse people who are not sure of
what they're doing.

On the other hand, maybe you're really asking about moving the
charge through an electric field, along a line where the gradient
is 12.5 volts per meter.  If that's the case, then I'll now go ahead
and answer the question that I just made up:

(12.5 volts/meter) x (2 meters) = 25 volts

25 volts means 25 joules/coulomb

(25 joule/coulomb) x (1 microcoulomb) = 25 microjoules

Finally, you didn't mention the sign of the charge (positive or negative
charge), and you didn't say whether it's moving in the direction parallel
to the field or opposite the field.  So we know that 25 microjoules of
energy passes between the charge and the field, but we don't know
in which direction.  We don't know if we have to push the charge
through the field, or if the field pushes the charge along. 
In other words, we don't know whether the charge is climbing or falling
through the field. 

You asked how much work "is required" to move the charge. It turns out
that if you use the right kind of charge and set up the field in the right
direction, you can GET work OUT of the charge as it moves.  In fact,
that's exactly how we run motors, light bulbs, talk on our smartphones,
and play our mp3s !

We can infer that star y is closer but other factors must be considered include the heat of the star, the mass, and the brightness of the star. even though star x seems dimmer it may just be less bright or contains less heat.