Considering Bohr's atomic model for hydrogen atom:
(A) the energy of H atom in ground state is same as energy of He$$^+$$ ion in its first excited state.
(B) the energy of H atom in ground state is same as that for Li$$^{++}$$ ion in its second excited state.
(C) the energy of H atom in ground state is same as that of He$$^+$$ ion for its ground state.
(D) the energy of He$$^+$$ ion in its first excited state is same as that for Li$$^{++}$$ ion in its ground state.
Choose the correct answer from the options below:

The energy of an electron in a Bohr orbit of any hydrogen-like species (single-electron ion) is given by
$$E_n = -\dfrac{13.6\,\text{eV}\,Z^{2}}{n^{2}}$$
where $$Z$$ is the atomic number and $$n = 1,2,3,\ldots$$ is the principal quantum number.

For the hydrogen atom $$\left(Z = 1\right)$$ in its ground state $$\left(n = 1\right)$$:
$$E_{\text{H, ground}} = -\dfrac{13.6 \cdot 1^{2}}{1^{2}} = -13.6\ \text{eV}$$

He$$^{+}$$ ion has $$Z = 2$$.
First excited state means $$n = 2$$.
$$E_{\text{He}^{+},\,n=2} = -\dfrac{13.6 \cdot 2^{2}}{2^{2}} = -13.6\ \text{eV}$$
This equals the hydrogen ground-state energy, so statement (A) is correct.

Li$$^{++}$$ ion has $$Z = 3$$.
Second excited state means $$n = 3$$.
$$E_{\text{Li}^{++},\,n=3} = -\dfrac{13.6 \cdot 3^{2}}{3^{2}} = -13.6\ \text{eV}$$
Again equal to the hydrogen ground-state energy, so statement (B) is correct.

He$$^{+}$$ ground state: $$n = 1$$.
$$E_{\text{He}^{+},\,n=1} = -\dfrac{13.6 \cdot 2^{2}}{1^{2}} = -54.4\ \text{eV}$$
This is not equal to $$-13.6\ \text{eV}$$, so statement (C) is incorrect.

He$$^{+}$$ first excited state energy has already been found as $$-13.6\ \text{eV}$$.
Li$$^{++}$$ ground state: $$n = 1$$.
$$E_{\text{Li}^{++},\,n=1} = -\dfrac{13.6 \cdot 3^{2}}{1^{2}} = -122.4\ \text{eV}$$
These are not equal, so statement (D) is incorrect.

Only statements (A) and (B) are correct. Hence the correct option is Option B: (A), (B) only.