Solution

The decay $\pi^0 \to \gamma + \gamma$, where the photons (and gamma quanta are photons) fly apart in opposite directions, has already been considered in problem 14.5.11. From there we take the formulas

$$E_2 = \frac{m_{\pi^0}^2 c^4}{4 E_1} \tag{1}$$
$$\mathcal{E}_K = E - m_{\pi^0} c^2 = E_1 + E_2 - m_{\pi^0} c^2 \tag{2}$$
Substituting (1) into (2):
$$\mathcal{E}_K = E_2 + \frac{m_{\pi^0}^2 c^4}{4 E_2} - m_{\pi^0} c^2 \tag{3}$$
where $E_1$ and $E_2$ are the energies of the forward‑ and backward‑going photons, respectively, and $\mathcal{E}_K$ is the desired kinetic energy of the $\pi^0$‑meson.

Now consider the second part of the condition. Collision with a heavy nucleus means that the photon momentum is almost entirely transferred to the nucleus. Owing to the large mass, the velocity acquired by the nucleus, and hence the energy spent on accelerating the nucleus (recall that it depends more strongly on velocity than on mass), are negligible. Then the threshold energy corresponds to the creation of particles at rest relative to each other:
$$E_2 = 2 m_e c^2. \tag{4}$$
Substituting (4) into (3), noting that the minimum $E_2$ corresponds to the minimum $\mathcal{E}_K$:
$$\mathcal{E}_{K \min} = 2 m_e c^2 + \frac{m_{\pi^0}^2 c^4}{8 m_e c^2} - m_{\pi^0} c^2 \tag{3'}$$
$$\mathcal{E}_{K \min} = 2 m_e c^2 \left( 1 + \left( \frac{m_{\pi^0} c^2}{4 m_e c^2} \right)^2 - \frac{m_{\pi^0} c^2}{2 m_e c^2} \right)$$
$$\mathcal{E}_K < 2 m_e c^2 \left( \frac{m_{\pi^0}}{4 m_e} - 1 \right)^2.$$