The kinetic energy of a moving particle is given by

$E=\frac{1}{2}m{u}^2 ...\left(1\right)$

When the final velocity of the particle becomes twice the initial velocity, the kinetic energy of the particle can be written as

$\begin{array}{rcl}nE& =& \frac{1}{2}m\left\{(2u{)}^2-u^2\right\}\\ & =& \frac{1}{2}m\left(3u^2\right) ...\left(2\right)\end{array}$

Divide equation (2) by equation (1) and solve to calculate the value of $n$.

$\begin{array}{rcl}\frac{nE}{E}& =& \frac{\frac{1}{2}m\left(3u^2\right)}{\frac{1}{2}m{u}^2}\\ & \Rightarrow & n=3\end{array}$