Answer
The magnitude of gravitational force, $F=1.5085\times10x^{-12}$ N and its direction is toward the y-axis as shown in figure.
Work Step by Step
Let's take a cartesian coordinate where x and y axes are as shown in the figure. Now let's take a very small piece of the rod $dl$ where $dl \to 0$ and the mass of the piece is $dm_1$. As the distance from the point $P$ and the piece is $R$ then the force for a single piece would be,
$$dF = G\frac{dm_{1}m_{2}}{R^2}$$
Here,
$G$ is the gravitational constant
$m_2$ is the mass of the object at point $P$
But as the rod is in mirror symetry in respect to the y-axis, the x-component of $dF$ would be cancel out by both side. So only the y-component will survive. So now we can write,
$$dF = G\frac{dm_{1}m_{2}}{r^2}cos(\theta)\;\; [only\;the\;y-component].....(eq. 1)$$
Now the mass of the rod for unit length is, $$\lambda = \frac{m_1}{\pi R}\\
So, \; dm_1 = \frac{m_1 dl}{\pi R}\\
or, \; dm_1 = \frac{m_1}{\pi R} R d\theta\\
or,\;dm_1= \frac{dm_1 d\theta}{\pi}$$
Now putting this to $eq.1$ we find,
$$dF = G\frac{m_1 m_2}{\pi R^2} sin\theta d\theta\\
or, F = G\frac{m_1 m_2}{\pi R^2} \int_{0}^\pi sin\theta d\theta\\
or, F = G\frac{m_1 m_2}{\pi R^2}[-cos\theta]_0^\pi\\
or,\;F=2G\frac{m_1 m_2}{\pi R^2}$$
Now putting values we can find that, $F = 1.5085 \times 10^-12 N\\$
And its direction is toward the y-axis.
