Answer
$$x=2.20\;m$$
Work Step by Step
Let $T_1$ is the tension in the cord which makes an angle $\theta=36.9^{\circ}$ and with the vertical and $T_2$ is the tension in the second cord which makes an angle $\phi=53.1^{\circ}$ and with the vertical. $W$ be weight of the bar.
At horizontal equilibrium: $T_1\sin\theta=T_2\sin\phi\;..........(1)$
At vertical equilibrium: $T_1\cos\theta+T_2\cos\phi=W\;.......(2)$
Under rotational equilibrium: $Wx=T_2\cos\phi L$
or, $x=\frac{T_2\cos\phi L}{W}$
or, $x=\frac{T_2\cos\phi L}{T_1\cos\theta+T_2\cos\phi}$ (using Eq. $2$)
or, $x=\frac{T_2\cos\phi L}{\frac{T_2\sin\phi}{\sin\theta}\cos\theta+T_2\cos\phi}$ (using Eq. $1$)
or, $x=\frac{\sin\theta\sin\phi L}{\sin\phi\cos\theta+\cos\phi\sin\theta}$
or, $x=\frac{\sin\theta\sin\phi L}{\sin(\theta+\phi)}$
Substituting the given values
$x=\frac{\sin36.9^{\circ}\times\sin53.1^{\circ}\times6.10}{\sin(36.9^{\circ}+53.1^{\circ})}\;m$
or, $\boxed{x=2.20\;m}$
