The total energy of the electron in the stationary states of the hydrogen atom is given by
$$
\mathrm{E}_{\mathrm{n}}=-\frac{m \mathrm{e}^4}{8 \mathrm{n}^2 \varepsilon_0^2 \mathrm{~h}^2}=(-13.6) \mathrm{eV} \quad[\text { if } \mathrm{n}=1]
$$
where signs are as usual and the $m$ that occurs in the Bohr formula is the reduced mass of electron and proton in hydrogen atom, the total energy of the electron in the ground state of the hydrogen atom is $-13.6 \mathrm{eV}$. For $\mathrm{H}-$ atom reduced mass $\mathrm{m}_{\mathrm{e}}$. Whereas for positronium, the reduced mass is
$$
\begin{aligned}
&\mathrm{m} \approx \frac{\mathrm{m}_{\mathrm{e}}}{2}, \text { by putting } \mathrm{m} \text { value in En, } \\
&\mathrm{E}_{\mathrm{n}}=\left(\frac{\left(\frac{-m e}{2}\right) \mathrm{e}^4}{8 \mathrm{n}^2 \varepsilon_0^2 \mathrm{~h}^2}\right)
\end{aligned}
$$
Hence, the total energy of the electron in the ground state of the positronium atom is
$$
\mathrm{E}_{\mathrm{n}}^{\prime}=\frac{-13.6 \mathrm{eV}}{2}=-6.8 \mathrm{eV}(\text { at }, \mathrm{n}=1)
$$