Solution

$\textbf{Method 1}$

Recall that for a photon the momentum is $p = \frac{E}{c}$.

The counter detected only one photon; therefore, the second one must have flown in the strictly opposite direction (if it had a component not parallel to the meson's direction of motion, there would be nothing to compensate it). Moreover, momentum conservation must hold:
$$p = \frac{E v}{c^2} = \frac{1}{c} (E_1 - E_2).$$
Energy conservation:
$$E = m_{\pi^0} c^2 + \mathcal{E}_K = E_1 + E_2.$$
Solve this system:
$$E_2 = E - E_1,$$
$$E v = c (2E_1 - E).$$
Denote $\beta = v/c$, write the definitions of energy and solve carefully:
$$\frac{m_{\pi^0} c^2 (1 + \beta)}{\sqrt{1 - \beta^2}} = 2E_1,$$
$$\sqrt{\frac{1 + \beta}{1 - \beta}} = \frac{2E_1}{m_{\pi^0} c^2} = s,$$
$$\beta = \frac{s^2 - 1}{s^2 + 1},$$
$$\mathcal{E}_K = E - m_{\pi^0} c^2 = m_{\pi^0} c^2 \left( \frac{1}{\sqrt{(1 + \beta)(1 - \beta)}} - 1 \right),$$
$$\mathcal{E}_K = m_{\pi^0} c^2 \left( \frac{s^2 + 1}{2s} - 1 \right),$$
$$\mathcal{E}_K = m_{\pi^0} c^2 \left( \frac{s}{2} + \frac{1}{2s} - 1 \right),$$
$$\mathcal{E}_K = m_{\pi^0} c^2 \left( \frac{E_1}{m_{\pi^0} c^2} + \frac{m_{\pi^0} c^2}{4E_1} - 1 \right).$$
Or, more compactly,
$$\mathcal{E}_K = \frac{(2E_1 - m_{\pi^0} c^2)^2}{4E_1} \approx 152\ \text{MeV}.$$
The formula in Savchenko's answer is incorrect, but the numerical answer is correct.

$\textbf{Method 2}$

The problem can also be solved by writing the invariant before and after decay:
$$I = \frac{E^2}{c^2} - p^2 = m_{\pi^0}^2 c^2,$$
$$I = \left( \frac{E_1 + E_2}{c} \right)^2 - \left( \frac{E_1 - E_2}{c} \right)^2 = \frac{4E_1 E_2}{c^2}.$$
Equating:
$$E_2 = \frac{m_{\pi^0}^2 c^4}{4E_1},$$
$$\mathcal{E}_K = E - m_{\pi^0} c^2 = E_1 + E_2 - m_{\pi^0} c^2,$$
$$\mathcal{E}_K = E_1 + \frac{m_{\pi^0}^2 c^4}{4E_1} - m_{\pi^0} c^2,$$
which leads to the same result.