Answer
$0.8c$
Work Step by Step
We know that the relativistic momentum is given by
$$p=\gamma \;mv$$
where $v$ is the speed of the particle, and $\gamma=\left[1-\frac{v^2}{c^2}\right]^{-1/2}$
Thus,
$$p=\dfrac{mv}{\left[1-\frac{v^2}{c^2}\right]^{1/2}}$$
Solving for $v$,
$$\left[1-\frac{v^2}{c^2}\right]^{1/2}=\dfrac{mv}{p}$$
Squaring both sides;
$$ 1-\frac{v^2}{c^2} =\dfrac{m^2v^2}{p^2}$$
$$ 1 =\dfrac{m^2v^2}{p^2}+\frac{v^2}{c^2}=v^2\left[\dfrac{m^2 }{p^2}+\frac{1}{c^2}\right]$$
Hence,
$$v=\sqrt{\dfrac{1}{\dfrac{m^2 }{p^2}+\frac{1}{c^2}}}$$
Plug the known;
$$v=\sqrt{\dfrac{1}{\dfrac{(1\times 10^{-3})^2 }{(400,000)^2}+\frac{1}{(3\times 10^8)^2}}}=\color{red}{\bf2 .4\times 10^8}\;\rm m/s$$
$$\boxed{v=0.8c}$$
