The radius of 2 ^ nd orbit of He ^ + of Bohr's model is r _ 1 and...

The formula to calculate the radius $\left(r\right)$ of any ${n}^{th}$ Bohr's orbit is given by

$r = {r}_{0}\dfrac{{n}^{2}}{Z} . . . \left(1\right)$

where, ${r}_{0}$ is Bohr's radius, $n$ is the principal quantum number and $Z$ is the atomic number.

Thus, for Helium ion,

$\begin{cases}{r}_{1} & = & {r}_{0}\dfrac{{2}^{2}}{2} \\ & = & 2{r}_{0} . . . \left(2\right)\end{cases}$

And, for the Beryllium ion,

$\begin{cases}{r}_{2} & = & {r}_{0}\dfrac{{4}^{2}}{4} \\ & = & 4{r}_{0} . . . \left(3\right)\end{cases}$

Divide equation (3) by equation (2) to obtain the required ratio.

$\begin{cases}\dfrac{{r}_{2}}{{r}_{1}} & = & \dfrac{4{r}_{0}}{2{r}_{0}} \\ & = & 2\end{cases}$

Hence, $x = 2$.