Step 1: Translate the wording.
The ground state is $n = 1$. First excited is $n = 2$, second is $n = 3$, and third excited is $n = 4$. So we work with $n = 4$.
Step 2: Recall the total energy formula.
For hydrogen, \[ E_n = -\frac{13.6}{n^2}\ \text{eV}. \]
Step 3: Find the total energy at $n = 4$.
\[ E_4 = -\frac{13.6}{16} = -0.85\ \text{eV}. \]
Step 4: Recall how potential energy links to total energy.
For the Coulomb attraction in an atom, the potential energy is exactly twice the total energy, \[ U_n = 2 E_n. \]
Step 5: Compute the potential energy.
\[ U_4 = 2 \times (-0.85) = -1.70\ \text{eV}. \]
Step 6: Sanity check with kinetic energy.
Kinetic energy is $+0.85$ eV, and adding it to $U = -1.70$ eV gives back $E = -0.85$ eV, which matches. \[ \boxed{U = -1.70\ \text{eV}} \]