C

Step 1: Identify the ground state energy of the H atom, which is given as -13.6 eV.
Step 2: Determine the second excited state energy. The first excited state energy is -3.4 eV, and the second excited state energy will be 0 eV (the ionization energy).
Step 3: The energy needed to ionize the H atom from the second excited state is 3.4 eV.
Therefore, the second excited state energy is -3.4 eV (as given in the question) and validates our calculation.
Hence the energy to ionize from the second excited state is the difference between its energy and the energy of free state.
Therefore, the answer is the value 3.4 eV.