The magnetic field due to the inner solenoid (within its volume) is

$B_1={\mu }_{0}n\left(2kt\right)$

The magnetic field due to the outer solenoid (within its volume) is

$B_2={\mu }_{0}n\left(kt\right)$

The flux of magnetic field through the circular area of radius $r$ is

$\varphi =B_1\pi R^2+B_2\pi r^2$

$\Rightarrow \varphi =\pi {\mu }_{0}nkt\left(2R^2+r^2\right)$

If $\overrightarrow{E}$ is the induced electric field at $r$, then

$\oint \overrightarrow{E}.d\overrightarrow{r}=-\frac{d\varphi }{dt}$

$\Rightarrow E\times 2\pi r=\pi {\mu }_{0}nk\left(2R^2+r^2\right)$

$\Rightarrow E={\mu }_{0}nk\left(\frac{R^2}{r}+\frac{r}{2}\right)$

The radius of the circle traversed by the charged particle is

$r=\frac{mv}{q{B}_2}$

$v=E\frac{qt}{m}={\mu }_{0}nk\left(\frac{R^2}{r}+\frac{r}{2}\right)\frac{qt}{m}$

$\Rightarrow {\mu }_{0}nk\left(\frac{R^2}{r}+\frac{r}{2}\right)\frac{qt}{m}=\frac{qr{B}_2}{m}$

$\Rightarrow {\mu }_{0}nkt\left(\frac{R^2}{r}+\frac{r}{2}\right)=r{\mu }_{0}nkt$

$\Rightarrow R^2+\frac{r^2}{2}=r^2$

$\Rightarrow r=R\sqrt{2}$