Answer
$133.5\;{\rm MeV}/ c^2$
Work Step by Step
Since the author assume that the initial kinetic energy of the given reaction is near to zero, so we can work with that assumption.
This means that the initial and final momentum of the system is zero.
This means that the two products have the same momentum magnitude.
$$p_{\pi^0}=p_{n}$$
Recall that $E^2=p^2c^2+m^2c^4$; So, $p^2c^2=E^2-m^2c^4 \tag 1 $ applying that by multiplying both sides by $c$ and then squaring both sides.
$$p_{\pi^0}^2\;c^2=p_{n}^2\;c^2$$
$$E_{\pi^0}^2-m_{\pi^0}^2c^4=E_{n}^2-m_{n}^2c^4$$
Solving for $m_{\pi^0}$;
$$m_{\pi^0}^2c^4=E_{\pi^0}^2-E_{n}^2+m_{n}^2c^4$$
$$m_{\pi^0} =\sqrt{\dfrac{E_{\pi^0}^2-E_{n}^2+m_{n}^2c^4}{c^4}}\tag 2$$
Now we need to find $E_{\pi^0}$ and $E_n$.
$$E_n=KE_n+m_nc^2\tag 3$$
Plugging the known;
$$E_n=(0.6)+(939.6)=\bf 940.2 \;\rm MeV\tag {3-a}$$
Now we need to find $E_{\pi^0}$.
We know that the energy is conserved, so
$$E_{\pi^-} +E_p =E_{\pi^0} +E_n $$
Thus;
$$m_{\pi^-}c^2+m_pc^2=E_{\pi^0}+E_n $$
Plugging $E_n$ from (3);
$$m_{\pi^-}c^2+m_pc^2=E_{\pi^0}+KE_n+m_nc^2$$
Hence;
$$E_{\pi^0}=m_{\pi^-}c^2+m_pc^2- m_nc^2-KE_n $$
Plugging the known;
$$E_{\pi^0}= 139.6+ 938.3- 939.6-0.6=\bf 137.7\;\rm MeV\tag 4$$
Now we need to plug (3-a) and (4) into (2);
$$m_{\pi^0} = \sqrt{137.7^2-940.2^2+ 939.6^2 } $$
$$m_{\pi^0} = \color{red}{\bf 133.5}\;{\rm MeV}/ c^2$$
which is close to its magnitude on table 32-2 (135 MeV/$c^2$)
