Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
According to the right-hand rule, the thumb finger is toward the current, the index finger is toward the magnetic field, and the middle finger is toward the force, the direction of the magnetic field is downward.
Now we have 3 forces exerted on the wire, the tension forces, the gravitational force, and the magnetic force, as shown below.
The $2T$ is due to the net tension in the two strings.
We can see that the rod is now at rest, so
$$\sum F_x=2T\sin\theta-F_B=0$$
Hence,
$$2T\sin\theta=BIL\tag 1$$
and
$$\sum F_y=2T\cos\theta-mg=0$$
Hence,
$$2T\cos\theta=mg$$
Thus,
$$T=\dfrac{mg}{2\cos\theta}$$
Plug into (1),
$$\dfrac{2mg}{2\cos\theta}\sin\theta=BIL $$
$$mg\tan\theta=BIL $$
Thus the strength of the magnetic field is given by
$$B=\dfrac{mg\tan\theta}{IL}$$
where $m=\mu L$ where $\mu$ here is the linear mass density of the rod.
$$B=\dfrac{\mu Lg\tan\theta}{IL}$$
$$\boxed{B=\dfrac{\mu g\tan\theta}{I }}$$
$$\color{blue}{\bf [b]}$$
We just need to plug the given into the boxed formula above,
$$ B=\dfrac{(0.055)(9.8)\tan12^\circ}{(10)}$$
$$B=\color{red}{\bf 11.5}\;\rm mT\tag{Downward}$$
