Answer
$v = 0.417~c$
Work Step by Step
We can write an expression for the rest energy:
$E = mc^2$
We can write an expression for the total energy:
$E = \gamma ~mc^2$
If the total energy is 10% more than the rest energy, then $\gamma = 1.1$
We can find the speed when $\gamma = 1.1$:
$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
$\sqrt{1-\frac{v^2}{c^2}} = \frac{1}{\gamma}$
$1-\frac{v^2}{c^2} = \frac{1}{\gamma^2}$
$\frac{v^2}{c^2} = 1-\frac{1}{\gamma^2}$
$v^2 = (1-\frac{1}{\gamma^2})~c^2$
$v = \sqrt{1-\frac{1}{\gamma^2}}~c$
$v = \sqrt{1-\frac{1}{1.1^2}}~c$
$v = \sqrt{0.1735537}~c$
$v = 0.417~c$