I presume that the rods are joined rigidly at O, not hinged. So they cannot rotate relative to each other, only as a single rigid body.

The force is applied to one rod, not both, and is distributed between them by internal forces which keep the structure rigid. In your attempt you have assumed that a force $F$ is applied to each rod, so that the total force on the structure is $2F$ instead of $F$.

Also you have assumed that the only forces on each rod are the applied force $F$ and the friction force on that rod. There are also forces and couples between the two rods which keep them aligned. If you analyse the forces on each rod using separate free body diagrams (FBDs) you must include forces which each rod exerts on the other. But it is simpler to model the whole structure as a single rigid body. Then you can ignore internal forces - ie those between the two rods.

There are 3 external forces acting horizontally on the whole structure :

1. applied force $F$ acting to the right at distance $x$ above point O

2. friction force $R_1=\mu mg$ acting to the left at distance $\frac12 L$ above O, and

3. friction force $R_2=4\mu mg = 4R_1$ acting to the left at distance $\frac12 L$ below O.

The rods move with constant velocity to the right and they remain perpendicular to the force $F$ - ie they do not turn. This means that the structure is in static equilibrium : in the frame of the COM it does not translate or rotate. So you can apply the two conditions for static equilibrium :

I. the resultant force on the structure is zero

II. the resultant moment of force on the structure about any point is zero.

Applying these conditions, taking moments about O, and substituting $R_2=4R_1$, we have

I. $F=R_1+R_2 = 5R_1$

II. $xF+\frac12 LR_2=\frac12 LR_1, xF=\frac12 LR_1-\frac42LR_1=-\frac32 LR_1$

Solving

$x=-\frac32 \frac{R_1}{F} L=-\frac{3}{10} L$

The force $F$ should be applied at a distance $\frac{3}{10} L$ **below** point O.

**Note:** Since this is a **statics** problem we might wonder why the question did not specify that the rods remain stationary, instead of moving with constant velocity.

When the rods move we can be sure that the friction forces on them are given by $R=\mu mg$. When the rods are stationary the friction forces can take any values up to $R=\mu mg$. The friction forces could be distributed in various ways and still fulfil the equilibrium conditions. See Is there a way to determine the exact value of static friction in this situation?

This means that the composite rod is in static equilibrium : in the frame of the COM it does not translate or rotate. So you can apply the TWO conditions for static equilibrium :

1. the resultant force on the object is zero

2, the resultant moment of force on the object is zero.