#### Answer

The particle's kinetic energy increases by a factor of 9

#### Work Step by Step

We can write an expression for the particle's initial kinetic energy when the particle's speed is $v$.
$KE_0 = \frac{1}{2}mv^2$
We can find the particle's kinetic energy when the speed is $3v$.
$KE = \frac{1}{2}m(3v)^2$
$KE = 9\times ~\frac{1}{2}mv^2$
$KE = 9\times ~KE_0$
The particle's kinetic energy increases by a factor of 9.