Answer
$\sqrt 2\;R$
Work Step by Step
We know that rotational energy is given by
$$K=\frac{1}{2}I\omega^2$$
So we need to double this energy without changing the angular velocity $\omega$. This means that we need to change the moment of inertia.
$$K_f = \frac{1}{2}I_f\omega^2 $$
Since $K_f=2K$, then
$$ \color{red}{\bf\not} \frac{1}{2}I_f\color{red}{\bf\not} \omega^2=2\times \color{red}{\bf\not} \frac{1}{2}I\color{red}{\bf\not} \omega^2$$
Thus,
$$I_f=2I\tag 1$$
This means that we need to double the moment of inertia of the given object.
And we also know that the moment of inertia is given by
$$I=\int r^2dm$$
which means that $I\propto r^2$; and for our case,
$$\dfrac{I}{I_f}=\dfrac{R^2}{R^2_f}$$
Plugging from (1);
$$\dfrac{\color{red}{\bf\not} I}{2\color{red}{\bf\not} I}=\dfrac{R^2}{R^2_f}$$
Hence,
$$R^2_f=2R$$
Thus,
$$\boxed{R_f=\sqrt2 R}$$
Therefore, $R$ must e increased by a factor of $\sqrt2$
