The given situation is shown in the figure



At highest point $A$,
$$
v_A=\sqrt{7 g r}
$$
$T_d$ and $T_s$ are tensions in the string when particle is at highest position $(A)$ and lowest position $(B)$.
$\therefore$ At point $A$.
$$
\begin{aligned}
& \frac{m v_A^2}{r}=T_A+m g \\
& \Rightarrow \quad \frac{m(\sqrt{7 g r})^2}{r}=T_A+m g \\
& \Rightarrow \quad 7 m g=T_A+m g \\
& \Rightarrow \quad T_A=6 \mathrm{mg} \\
&
\end{aligned}
$$
At point $B$ by the law of conservation of energy, change in potential energy $=$ Change in kinetic energy
$$
\begin{aligned}
& m g(2 r)=\frac{1}{2} m v_{\bar{B}}^2-\frac{1}{2} m v_A^2 \\
\Rightarrow \quad & 2 m g r=\frac{1}{2} m v_B^2-\frac{1}{2} m(7 g r) \\
\Rightarrow \quad & 4 g r=v_B^2-7 g r \\
\Rightarrow \quad & v_s=\sqrt{11 g r}
\end{aligned}
$$
At lowest point $B, T_n-\frac{m v_n^2}{r}=m g$
$$
\begin{aligned}
T_s-\frac{m}{r} \cdot 11 g r & =m g \\
\Rightarrow \quad T_n-11 m g & =m g \\
\Rightarrow \quad T_s & =12 m g \\
\frac{T_A}{T_s} & =\frac{6 m g}{12 m g}=\frac{1}{2} \\
\Rightarrow \quad T_A: T_s & =1: 2
\end{aligned}
$$