The system consists of two masses $m$ connected by a light rod of length $L$. Total mass $M = 2m$. A constant external force $F$ acts on one ball.
Acceleration of CM: $$ a_{cm} = \frac{F_{ext}}{M} = \frac{F}{2m} $$ The CM moves in a straight line along the direction of F (East).

The force $F$ creates a torque about the Center of Mass. Let $\theta$ be the angle between the rod and the force vector $F$. Initially, the rod is aligned along a longitude (North-South) and $F$ is East, so initially $\theta = 90^\circ$.
Torque about CM: $$ \tau = F \cdot \frac{L}{2} \sin\theta $$ Moment of Inertia about CM: $$ I = m\left(\frac{L}{2}\right)^2 + m\left(\frac{L}{2}\right)^2 = \frac{mL^2}{2} $$

As the force pulls, the rod rotates and aligns with the force. Using the Work-Energy theorem in the CM frame: $$ \text{Work done by Torque} = \Delta KE_{rot} $$ $$ \int_{90^\circ}^{\theta} \tau \, d\theta_{rot} $$ Note: The angle $\theta$ decreases from $90^\circ$ to $0$. The work done by torque as the rod swings from perpendicular to angle $\theta$ is: $$ \frac{1}{2} I \omega^2 = \int_{\theta}^{\pi/2} F \frac{L}{2} \sin\phi \, d\phi $$ $$ \frac{1}{2} \left(\frac{mL^2}{2}\right) \omega^2 = \frac{FL}{2} [-\cos\phi]_{\theta}^{\pi/2} = \frac{FL}{2} \cos\theta $$ $$ \frac{mL^2}{4} \omega^2 = \frac{FL}{2} \cos\theta \implies m \omega^2 \frac{L}{2} = F \cos\theta $$

Consider the ball without the external force F. The only force acting on it is the Tension $T$ directed along the rod towards the CM.
In the inertial frame, this mass has acceleration $\vec{a} = \vec{a}_{cm} + \vec{a}_{rad} + \vec{a}_{tan}$. Along the rod (radial direction), the equation of motion is: $$ T = m(a_{cm} \cos\theta + \omega^2 \frac{L}{2}) $$ (Here $a_{cm} \cos\theta$ is the component of linear acceleration along the rod).

Substitute $a_{cm} = \frac{F}{2m}$ and $m \omega^2 \frac{L}{2} = F \cos\theta$: $$ T = m\left( \frac{F}{2m} \cos\theta \right) + F \cos\theta $$ $$ T = \frac{F}{2} \cos\theta + F \cos\theta = 1.5 F \cos\theta $$

The tension is maximum when $\cos\theta$ is maximum, i.e., when $\theta = 0^\circ$ (rod aligns with the force). $$ T_{max} = 1.5 F (1) = 1.5 F $$