Understand the composition of a hydrogen atom and Positronium.

A hydrogen atom consists of a proton (mass $m_p$, charge $+e$) and an electron (mass $m_e$, charge $-e$). Its ground state energy is $E_H = -13.6 \text{ eV}$.
Positronium consists of a positron (mass $m_e$, charge $+e$) and an electron (mass $m_e$, charge $-e$).

The reduced mass $\mu$ for a two-particle system with masses $m_1$ and $m_2$ is given by $\mu = \frac{m_1 m_2}{m_1 + m_2}$.
For a hydrogen atom, $m_1 = m_p$ and $m_2 = m_e$.

Since the mass of a proton ($m_p$) is much greater than the mass of an electron ($m_e$), $m_p + m_e \approx m_p$.
Therefore, $\mu_H \approx \frac{m_p m_e}{m_p} = m_e$.

For Positronium, $m_1 = m_e$ (positron) and $m_2 = m_e$ (electron).

The energy levels of a hydrogen-like atom are directly proportional to the reduced mass of the system.

Therefore, the ratio of the ground state energy of Positronium to that of hydrogen is:

Using the approximate values for reduced masses:

So, $E_{Ps} = \frac{1}{2} E_H$.
Given $E_H = -13.6 \text{ eV}$ for the ground state of hydrogen:

The more precise calculation using $\mu_H = \frac{m_p m_e}{m_p + m_e}$ yields:

Since $m_p \approx 1836 m_e$:

This value is very close to 0.5, so the approximation holds well.

Rounding to one decimal place, this is $-6.8 \text{ eV}$.

The correct option is (c).