Answer
The angle the rod makes with the horizontal is $7^o$
Work Step by Step
The rod is under 3 forces: the rod's weight $W$ pointing downward, 2 restoring forces from 2 springs $F_1$ and $F_2$ pointing downward $(k_1=59N/m, k_2=33N/m)$. Because the system is in equilibrium, these forces balance each other. $$F_1+F_2=W$$ $$k_1x_1+k_2x_2=1.4\times9.8$$ $$59x_1+33x_2=13.72 (1)$$
If we take the axis of rotation to pass through the center of the rod, where $W$ is located, then $W$ produces no torque.
$F_1$ and $F_2$, on the other hand, both produce torques in opposite directions with lever arm being half the length of the rod, which is $0.375m$
The net torque needs to be zero, so $$0.375(F_1-F_2)=0$$ $$F_1-F_2=0$$ $$59x_1-33x_2=0 (2)$$
Solve (1) and (2), we have $x_1=0.116m$ and $x_2=0.208m$
Take $\theta$ to be the angle the rod makes with the horizontal and $l$ to be the rod's length: $$\tan\theta=\frac{x_2-x_1}{l}=\frac{0.208m-0.116m}{0.75m}=0.123$$ $$\theta=7^o$$
