Answer
${\bf 0.845}c$
Work Step by Step
According to the conservation of energy principle, the energy of the electron plus the energy of the positron plus the energy of the two ejected gamma photons must equal the energy of the proton and the antiproton before the collision.
Thus,
$$2E_e+2E_\gamma=2E_p $$
Hence,
where $e$ is for the electron (the mass of the electron and the position is the same and they are ejected at the same speed), $\gamma$ for the two ray photons, $p$ is for the proton (the mass of the proton and the antiproton is the same and they collide at the same speed).
$$E_p=E_e+E_\gamma\tag 1 $$
We need to use the relativistic energy equation to find the energies.
$$E_e=\gamma\; m_ec^2$$
where $\gamma= \left[1-\dfrac{u^2}{c^2}\right]^{-1/2}$ where $u_e=0.9999995c$, so $\gamma=\bf 1000.000$,
Hence,
$$E_e=1000m_ec^2\tag 2$$
and the energy of the two ray photons,
$$E_\gamma=\dfrac{hc}{\lambda}\tag 3$$
Plug (2) and (3) into (1),
$$E_p=1000m_ec^2+\dfrac{hc}{\lambda} $$
And since we need to find the speed before the collision, $E_p=\gamma_pm_pc^2$ where $\gamma_p= \left[1-\dfrac{u_p^2}{c^2}\right]^{-1/2}$ where $u_p$ is the speed of the proton.
$$\left[1-\dfrac{u_p^2}{c^2}\right]^{-1/2}m_pc^2=1000m_ec^2+\dfrac{hc}{\lambda} $$
$$\left[1-\dfrac{u_p^2}{c^2}\right]^{-1/2} =\dfrac{1000m_ec +\dfrac{h }{\lambda} }{m_pc}$$
$$\left[1-\dfrac{u_p^2}{c^2}\right]^{ 1/2} =\dfrac{m_pc}{1000m_ec +\dfrac{h }{\lambda} }$$
Squaring both sides;
$$ 1-\dfrac{u_p^2}{c^2} =\left[\dfrac{m_pc}{1000m_ec +\dfrac{h }{\lambda} }\right]^2$$
$$ u_p^2 =c^2\left(1-\left[\dfrac{m_pc}{1000m_ec +\dfrac{h }{\lambda} }\right]^2\right)$$
$$ u_p =c \left(\sqrt{1-\left[\dfrac{m_pc}{1000m_ec +\dfrac{h }{\lambda} }\right]^2}\right)$$
Plug the known;
$$ u_p =c \left(\sqrt{1-\left[\dfrac{(1.67\times 10^{-27})(3\times 10^8)}{1000(9.11\times 10^{-31})(3\times 10^8) +\dfrac{6.63\times 10^{-34}}{(1\times 10^{-6}\times 10^{-9})} }\right]^2}\right)$$
$$u_p=\color{red}{\bf 0.845}c$$
