The energy levels of a hydrogen atom are given by the formula :

$$E_n = -\frac{13.6 ~\mathrm{eV}}{n^2}$$

where $E_n$ is the energy of the $n$-th energy level.

The ground state of hydrogen ($n=1$) has an energy of $-13.6 ~\mathrm{eV}$ as mentioned, which means that it would take $+13.6 ~\mathrm{eV}$ to ionize it (remove the electron completely) from this state, since ionization implies moving the electron to a state of zero energy.

The second excited state of hydrogen is when $n=3$ (as $n=1$ is the ground state and $n=2$ is the first excited state). Thus, the energy of the second excited state is :

$$E_3 = -\frac{13.6 ~\mathrm{eV}}{3^2} = -\frac{13.6 ~\mathrm{eV}}{9} = -1.51 ~\mathrm{eV}$$

Since ionization implies moving the electron from its current energy level to $0$ energy, the energy required to ionize the atom from this state is the absolute value of its current energy state. So, it will take $+1.51 ~\mathrm{eV}$ to ionize a hydrogen atom from its second excited state.

So, the correct answer is Option C : $1.51 ~\mathrm{eV}$.