A square with side length s = 2a has four charges at its corners: three charges of +Q and one charge of -Q. What is the electric potential at the center of the square?

The side length of the square is $s = 2a$.
Point P is the center of the square. The distance from any corner to the center of a square is half the length of its diagonal.
The length of the diagonal $d$ of a square with side length $s$ is $d = s\sqrt{2}$.
In this case, $d = (2a)\sqrt{2}$.
The distance $r$ from each charge to point P is half the diagonal:

$$r = \frac{d}{2} = \frac{(2a)\sqrt{2}}{2} = a\sqrt{2}$$

The electric potential $V$ due to a point charge $q$ at a distance $r$ is given by the formula:

$$V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$$

There are three charges of $+Q$ and one charge of $-Q$. All are at the same distance $r = a\sqrt{2}$ from point P.

Let $V_1, V_2, V_3$ be the potentials due to the three $+Q$ charges, and $V_4$ be the potential due to the $-Q$ charge.

$$V_1 = \frac{1}{4\pi\epsilon_0} \frac{+Q}{a\sqrt{2}}$$ $$V_2 = \frac{1}{4\pi\epsilon_0} \frac{+Q}{a\sqrt{2}}$$ $$V_3 = \frac{1}{4\pi\epsilon_0} \frac{+Q}{a\sqrt{2}}$$ $$V_4 = \frac{1}{4\pi\epsilon_0} \frac{-Q}{a\sqrt{2}}$$

The total potential $V_{total}$ at P is the algebraic sum of the individual potentials:

$$V_{total} = V_1 + V_2 + V_3 + V_4$$ $$V_{total} = \frac{1}{4\pi\epsilon_0} \frac{Q}{a\sqrt{2}} + \frac{1}{4\pi\epsilon_0} \frac{Q}{a\sqrt{2}} + \frac{1}{4\pi\epsilon_0} \frac{Q}{a\sqrt{2}} + \frac{1}{4\pi\epsilon_0} \frac{-Q}{a\sqrt{2}}$$

Factor out the common term $\frac{1}{4\pi\epsilon_0 a\sqrt{2}}$:

$$V_{total} = \frac{1}{4\pi\epsilon_0 a\sqrt{2}} (Q + Q + Q - Q)$$ $$V_{total} = \frac{1}{4\pi\epsilon_0 a\sqrt{2}} (2Q)$$ $$V_{total} = \frac{2Q}{4\pi\epsilon_0 a\sqrt{2}}$$

$$V_{total} = \frac{Q}{2\pi\epsilon_0 a\sqrt{2}}$$

This can also be written as:

$$V_{total} = \frac{Q}{2\sqrt{2}\pi\epsilon_0 a}$$

Comparing this result with the given options:
A) $\frac{Q}{2\sqrt{2}\pi\epsilon_0 a}$
B) $\frac{Q}{\sqrt{2}\pi\epsilon_0 a}$
C) $\frac{Q}{2\pi\epsilon_0 a}$
D) $\frac{Q}{4\pi\epsilon_0 a}$

The calculated total potential matches option A.

The final answer is $\boxed{\text{A}}$