Minimum Force

First of all we need to understand that for the middle rod to be horizontal the only way is that which is shown in the figure. For any other case it won't be horizontal (as the supports are rods and not strings and hence can't be compressed.)
Now, let us assume that the force $F$ is acted on hinge $D$ making an angle $x$ with the horizontal. $T_{1}$ , $T_{2}$ and $T_{3}$ are the tensions in rod $A$ $C$ , $C$ $D$ and $D$ $B$ respectively.
Now, from vertical equilibrium of hinge $C$ , we have,
$T_{1}$ $\dfrac{\sqrt{3}}{2}$ $=$ $m$ $g$ $....... (i)$
From horizontal equilibrium of hinge $C$ , we have,
$\dfrac{1}{2}$ $T_{1}$ $=$ $T_{2}$ $....... (ii)$
Again, from horizontal equilibrium of hinge $D$ , we have,
$\dfrac{1}{2}$ $T_{3}$ $+$ $F$ $\cos x$ $=$ $T_{2}$ $....... (iii)$
Finally, from vertical equilibrium of hinge $D$ , we have,
$F$ $\sin x$ $=$ $T_{3}$ $\dfrac{\sqrt{3}}{2}$ $....... (iv)$
Solving the above equations we get, $F$ $=$ $\dfrac{mg}{\sin x + \sqrt{3}\cos x}$

Now, for minimum $F$ , the denominator i.e. $(\sin x + \sqrt{3}\cos x)$ should be maximum i.e. $2$ .

Therefore, we have, $F$ $=$ $\dfrac{mg}{2}$

Hence, $\boxed{k = 2.00}$

As all the rods are weightless, the external force T1 in the hinge A is directed along side AC. Similarly the external force T2 in the hinge B is directed along side BD. Let the point of intersection of these T1 and T2 forces be K. In the equilibrium the corresponding sum of the torques of all external forces T1, T2, mg and F (with the axis at point K) should be equal to zero. Then force mg multiplied by its lever arm l/2 (its torque according to point K) should be equal to the corresponding torque of the force F. It is easily seen that the force F will be minimum when its lever arm has maximum value (i.e. is equal to l). Hence F = mg/2.