$X_C=\dfrac{1}{\omega C}$, so $X_C$ vs $(1/\omega)$ is a straight line through the origin with slope $=1/C$.

From the given graph, the capacitor with the smaller slope has the LARGER capacitance (since slope $\propto 1/C$).

Method: read the slopes $m_1$ (for $C_1$) and $m_2$ (for $C_2$) off the graph; then $C_1/C_2 = m_2/m_1$ (inverse ratio of slopes). (The exact numeric slope values were not legible in the extracted question — apply this relation to the actual graph.)