Step 1: Write the energy levels.
For hydrogen, the level energy is \[ E_n = -\frac{13.6}{n^2}\ \text{eV}, \] with ground state $n = 1$ giving $E_1 = -13.6\ \text{eV}$.
Step 2: Identify the second excited state.
The ground state is $n=1$, the first excited state is $n=2$, so the second excited state is $n = 3$.
Step 3: Energy of the $n=3$ level.
\[ E_3 = -\frac{13.6}{9} \approx -1.51\ \text{eV}. \]
Step 4: Excitation energy to $n=3$.
\[ \Delta E = E_3 - E_1 = (-1.51) - (-13.6) = 12.09 \approx 12\ \text{eV}. \]
Step 5: Ionization energy.
Ionization takes the electron to $n=\infty$ where $E_\infty = 0$, so \[ \Delta E_{\text{ion}} = 0 - (-13.6) = 13.6\ \text{eV}. \]
Step 6: Conclude.
The energy to reach the second excited state is about $12\ \text{eV}$, and to ionize is $13.6\ \text{eV}$.
\[ \boxed{(a)\ \sim 12\ \text{eV},\ (b)\ 13.6\ \text{eV}} \]