I setup the problem using energy conservation: K_a + U_a = K_b + U_b:

K_a = total kinetic energy of the 3 charges when they're at rest
U_a = total potential energy of the 3 charges when they're at rest
K_b = total kinetic energy of the 3 charges when they're far apart
U_b = total potential energy of the 3 charges when they're far apart

K_a = 0 since the charges don't initially move
U_b = 0 when they're far away (when they're infinitely far away)

That leaves: U_a = K_b
where U_a = 3*(k*q_1*q_2)/r so
K_b = 3*(k*q_1*q_2)/r
and the answer I got was 3 * (8.9*10^9 * (1.6*10^-19)^2)/(8.50×10^-10) = 8.04*10^-19 which is incorrect.
Am I on the right track solving the problem? What am I overlooking?

Your error was assuming that you simply triple the amount of energy of a single pair of charges.

Do it this way: place the first proton in empty space, how much work to do it? none, since no other charges are pushing it away.

Bring in the second proton. HOw much work? THe same amount that is equal to the potential energy when they are in place: (kq^2)/r

Now, when the alpha is brought in, it is being pushed away by two charges that are already there. The spot that the third charge (the alpha) goes into has a potential (voltage) that is the algebraic (not vector) sum of the potentials provided by the two protons in place. use the equation V=kq/r twice, once for each proton, to find the potential for that spot, then multiply the charge of the alpha to the potential to find potential energy. Add this to the potential energy of the two protons you already had for total energy.

Let me see if I get this right, the potential to place the first proton is 0 so the work or potential energy required is 0.

The potential to place the second proton, due to the first proton, is (k*q)/r and the potential energy is (k*q^2)/r.

The potential to place the alpha particle is the sum of the potentials of both protons, (k*q)/r + (k*q)/r or 2(k*q)/r and the potential energy required is: q_alpha * 2(k*q)/r.

And the total energy is the sum of the potential energy of the system? 0 + (k*q^2)/r + q_alpha * 2(k*q)/r
Am I also correct that the alpha also has charge 1.6*10^-19? What about when the particles move very far away (towards infinity), would the total energy be 0?

The alpha particle has two protons and two nutrons, so its charge is twice that of the proton. Otherwise, your summary is correct.

When the particles move to "infinity," we assume that no other object has interfered with them, and all the total energy is still there in the form of KE. Inertia rules!