1. Use the energy formula for hydrogen atom

   The energy of the electron in the $n$th orbit of hydrogen is $E_n = \frac{-13.6}{n^2}\,\text{eV}$

2. Identify the 2nd excited state

   • Ground state $\to n=1$
   • 1st excited state $\to n=2$
   • 2nd excited state $\to n=3$

   So we need the energy at $n=3$.

3. Calculate the energy

   $E_3 = \frac{-13.6}{3^2} = \frac{-13.6}{9}$

   $E_3 \approx -1.51\,\text{eV}$

4. Compare with options

   • A: $-1.51\,\text{eV}$ ✅
   • B: $-3.4\,\text{eV}$
   • C: $-6.04\,\text{eV}$
   • D: $-13.6\,\text{eV}$

Therefore, the correct answer is A.