Answer
$v = 0.943~c$
Work Step by Step
We can write an expression for the rest energy:
$E = mc^2$
We can write an expression for the kinetic energy:
$E = (\gamma-1) ~mc^2$
If the kinetic energy is twice the rest energy, then: $~~\gamma - 1 = 2$
We can find the speed when $\gamma = 3$:
$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
$\sqrt{1-\frac{v^2}{c^2}} = \frac{1}{\gamma}$
$1-\frac{v^2}{c^2} = \frac{1}{\gamma^2}$
$\frac{v^2}{c^2} = 1-\frac{1}{\gamma^2}$
$v^2 = (1-\frac{1}{\gamma^2})~c^2$
$v = \sqrt{1-\frac{1}{\gamma^2}}~c$
$v = \sqrt{1-\frac{1}{3^2}}~c$
$v = \sqrt{0.8889}~c$
$v = 0.943~c$