Both the orbital size and the level energy in a hydrogen-like two-body system are governed by the reduced mass $\mu$, so the whole comparison reduces to comparing $\mu$ for the two atoms.
For ordinary hydrogen the nucleus (proton) is about 1836 times heavier than the electron, so the reduced mass is essentially the electron mass, $\mu_H \approx m_e$.
Positronium has two particles of equal mass $m_e$, so its reduced mass is exactly half the electron mass: $\mu_p = m_e/2$. This is the single fact that drives everything.
Radius: the Bohr radius is inversely proportional to reduced mass, $r \propto 1/\mu$. Halving $\mu$ doubles $r$, so $r_p = 2 r_H$.
Energy: the level energy is directly proportional to reduced mass, $E \propto \mu$. Halving $\mu$ halves the energy magnitude, so $E_p = E_H/2$.
Both conditions match option (A).
\[\boxed{r_p = 2r_H \text{ and } E_p = E_H/2}\]