In a hypothetical hydrogen-like atom, the electron is replaced by a...

According to Bohr's quantization condition, the angular momentum $L$ in the $n^{\text{th}}$ orbit is given by $L = \dfrac{nh}{2\pi}$.

Given $L = \dfrac{3h}{\pi}$, we have $\dfrac{nh}{2\pi} = \dfrac{3h}{\pi} \Rightarrow n = 6$.

The radius of the $n^{\text{th}}$ Bohr orbit is inversely proportional to the mass of the orbiting particle and directly proportional to $n^2$ ($r \propto \dfrac{n^2}{m}$).
For the normal hydrogen atom, $r_1 = a_0$.
For the hypothetical atom, $r_6 = a_0 \times \dfrac{6^2}{5} = \dfrac{36}{5} a_0 = 7.2 a_0$.

The energy of the $n^{\text{th}}$ Bohr orbit is directly proportional to the mass of the orbiting particle and inversely proportional to $n^2$ ($E \propto \dfrac{m}{n^2}$).
For the normal hydrogen atom, $E_1 = E_0$.
For the hypothetical atom, $E_6 = E_0 \times \dfrac{5}{6^2} = \dfrac{5}{36} E_0$.

Answer: $7.2 a_0, \dfrac{5 E_0}{36}$