Answer
$$\frac{T_A}{T_B}=\frac{1}{27}$$
Work Step by Step
For a satellite orbiting around the Earth (mass $M$), its speed is given by $$v=\sqrt{\frac{GM}{r}}=\frac{(GM)^{1/2}}{r^{1/2}}$$ $$r^{1/2}=\frac{(GM)^{1/2}}{v}$$ $$r^{3/2}=\frac{(GM)^{3/2}}{v^3} (1)$$
The tension of the satellite is given by $$T=\frac{2\pi r^{3/2}}{(GM)^{1/2}}$$
Plug (1) here: $$T=\frac{2\pi}{(GM)^{1/2}}\times\frac{(GM)^{3/2}}{v^3}=\frac{2\pi GM}{v^3}$$
Now we calculate $\frac{T_A}{T_B}$ $$\frac{T_A}{T_B}=\frac{\frac{2\pi GM}{v_A^3}}{\frac{2\pi GM}{v_B^3}}=\frac{v_B^3}{v_A^3}$$
We know that $v_A=3v_B$. Therefore, $$\frac{T_A}{T_B}=\Big(\frac{1}{3}\Big)^3=\frac{1}{27}$$
