Answer
$9:4$
Work Step by Step
We know that the spring constant is given by
$$T=2\pi \sqrt{\dfrac{m}{k}}$$
Hence,
$$T^2=4\pi^2\left[ \dfrac{m}{k} \right]$$
So,
$$k= \dfrac{4\pi^2m}{T^2} $$
Now we need to find the ratio of $k_A/k_B$
$$\dfrac{k_A}{k_B}=\dfrac{\dfrac{4\pi^2m_A}{T_A^2}}{\dfrac{4\pi^2m_B}{T_B^2}}=\dfrac{ \color{red}{\bf\not}4 \color{red}{\bf\not}\pi^2 \color{red}{\bf\not}m_A}{T_A^2}\cdot\dfrac{T_B^2} { \color{red}{\bf\not}4 \color{red}{\bf\not}\pi^2 \color{red}{\bf\not}m_B}$$
Recall that the two systems have the same mass, so $m_A=m_B$.
Thus,
$$\dfrac{k_A}{k_B}=\left(\dfrac{T_B}{T_A}\right)^2 $$
from the given graph, we can see that $T_B=6\;\rm s$ and that $T_A=\frac{12}{3}=4\;\rm s$.
Therefore,
$$\dfrac{k_A}{k_B}=\left(\dfrac{6}{4}\right)^2 =\dfrac{9}{4}=\color{red}{\bf9:4 }$$
