Answer
$\beta = 0.94107519$
Work Step by Step
We can find $\gamma$:
$K = (\gamma-1)~mc^2$
$\gamma-1 = \frac{K}{mc^2}$
$\gamma = 1+\frac{K}{mc^2}$
$\gamma = 1+\frac{(1.0000000\times 10^6~eV)(1.602 176 462\times 10^{-19}~J/eV)}{(9.109 381 88\times 10^{-31}~kg)(2.998\times 10^8~m/s)^2}$
$\gamma = 1+ 1.9568529$
$\gamma = 2.9568529$
We can find $\beta$:
$\gamma = \frac{1}{\sqrt{1-\beta^2}}$
$1-\beta^2 = \frac{1}{\gamma^2}$
$\beta^2 = 1-\frac{1}{\gamma^2}$
$\beta = \sqrt{1-\frac{1}{\gamma^2}}$
$\beta = \sqrt{1-\frac{1}{(2.9568529)^2}}$
$\beta = 0.94107519$
