Answer
In the first case, the distances are equal: $1.3\;\rm mm$
In the first case, the distances are also equal: $0.65\;\rm mm$
Work Step by Step
We know that each lie formed by an ion is at a distance of $2r$ from $\rm S_2$; whereas $r$ is the radius of the curvature.
So, we need to find the radius of the curvature of each ion and after that, we can find the distances between the formed lines.
We know that the radius of curvature is given by
$$m=\dfrac{qBB^{'}r}{E}$$
Thus,
$$r=\dfrac{mE}{q BB^{'}}=\dfrac{mE}{q B^2}$$
Noting that the two magnetic fields are equal.
And hence the distance from $\rm S_2$ is given by
$$ x=2r=\dfrac{2mE}{q B^2}\tag 1$$
Thus, for the first ion (carbon isotope of mass number 12):
$$ x_{isotope-12} =\dfrac{2m_{isotope-12}E}{q BB^{'}}=\dfrac{2\cdot 12\cdot(1.67\times10^{-27})\cdot 2.88\times 10^4}{(1.6\times 10^{-19})(0.68^2)}$$
$$x_1=\color{blue}{\bf15.6}\;\rm mm$$
$$ x_{isotope-13} =\dfrac{2m_{isotope-13}E}{q BB^{'}}=\dfrac{2\cdot 13\cdot(1.67\times10^{-27})\cdot 2.88\times 10^4}{(1.6\times 10^{-19})(0.68^2)}$$
$$x_2=\color{blue}{\bf 16.9}\;\rm mm$$
$$ x_{isotope-14} =\dfrac{2m_{isotope-14}E}{q BB^{'}}=\dfrac{2\cdot 14\cdot(1.67\times10^{-27})\cdot 2.88\times 10^4}{(1.6\times 10^{-19})(0.68^2)}$$
$$x_3=\color{blue}{\bf 18.2}\;\rm mm$$
Hence, the distances between the lines are given by
$$\Delta x_{12}=x_2-x_1=16.9-15.6$$
$$\Delta x_{12}=\color{red}{\bf 1.3}\;\rm mm$$
$$\Delta x_{23}=x_3-x_2=18.2-16.9$$
$$\Delta x_{23}=\color{red}{\bf 1.3}\;\rm mm$$
$\text{When the charge is doubled:}$
The charge in (1) is on the denominator, so doubling it halved the distance.
So,
$$x_1=\color{blue}{\bf 7.8}\;\rm mm$$
$$x_2=\color{blue}{\bf 8.45}\;\rm mm$$
$$x_3=\color{blue}{\bf 9.1}\;\rm mm$$
Hence, the new distances between the lines are given by
$$\Delta x_{12}=x_2-x_1=8.45-7.8$$
$$\Delta x_{12}=\color{red}{\bf 0.65}\;\rm mm$$
$$\Delta x_{23}=x_3-x_2=9.1-8.45$$
$$\Delta x_{23}=\color{red}{\bf 0.65}\;\rm mm$$
