How to find the velocity of point of intersection? 
My approach: The velocity of the point of intersection u will be in horizontal direction due to symmetry about x-axis.The velocity v of the rod makes angle $\theta$ with the vertical.Thus, $v\sin \theta=u$.
But the correct answer is A. Can anyone please point out where am I wrong?
 A: TL;DR You have correctly identified that the intersection point does not travel along the vertical axis, but your conclusion about the horizontal velocity is not correct. I show here how to solve this problem by following: (i) geometric approach, from which you will see what is the problem with your conclusion, and (ii) algebraic approach, which is in my opinion much more straightforward for this particular problem.

Geometric approach
Let rod $A$ be the one with negative slope. If the positive horizontal axis goes along the dashed line to the right, then velocity of the rod $A$ is
$$\vec{v}_{A/0} = v \angle \bar{\theta} = v ( \sin\theta \hat{\imath} + \cos\theta \hat{\jmath})$$
where $\bar{\theta} = 90^\circ - \theta$. Here you assumed that the intersection horizontal velocity is simply $v \sin\theta$, which is incorrect. That is horizontal velocity of a point (particle) on the rod, which is not the same as the intersection. See figure below for geometric explanation.

Figure: Geometric approach to calculate intersection velocity. The two red lines are of same length.
The intersection horizontal displacement $\Delta x$ can be calculated as
$$\frac{\Delta x}{\sin 90^\circ} = \frac{v \Delta t}{\sin \theta} \qquad \rightarrow \qquad \boxed{\frac{\Delta x}{\Delta t} = \frac{v}{\sin\theta} = v \cdot \mathrm{cosec}(\theta)}$$
where $\Delta x / \Delta t$ equals intersection velocity.

Algebraic approach
Let's describe the two rods as lines and place them in a (fixed) Cartesian coordinate system. The line equations are then
$$y_1 = x \tan\theta_1 \qquad y_2 = -x \tan\theta_2$$
The lines travel at certain velocity, but their slope remains constant. This means that the line offset changes in time (see figure below). It takes only a little bit of geometry to show that the offset magnitude is $vt/\cos\theta$ and the line equations become
$$y_1(t) = x(t) \tan\theta_1 - \frac{v_1 t}{\cos\theta_1} \qquad \text{and} \qquad y_2(t) = -x(t) \tan\theta_2 + \frac{v_2 t}{\cos\theta_2}$$
Your problem is actually a simple version with $\theta_1 = \theta_2 \equiv \theta$ and $v_1 = v_2 \equiv v$
$$y_1(t) = x(t) \tan\theta - \frac{v t}{\cos\theta} \qquad \text{and} \qquad y_2(t) = -x(t) \tan\theta + \frac{v t}{\cos\theta}$$

Figure: Algebraic approach to calculate intersection velocity. Line offset changes in time.
Coordinates of the intersection are found from the condition $y_1(t) = y_2(t)$
$$x_i(t) = \frac{v t}{\sin\theta} \qquad \text{and} \qquad y_i(t) = 0$$
The intersection velocity components are
$$v_{i,x} = \frac{d}{dt} x_i(t) = \frac{v}{\sin\theta} = v \cdot \mathrm{cosec}(\theta) \qquad \text{and} \qquad v_{i,y} = \frac{d}{dt} y_i(t) = 0$$
and the intersection velocity magnitude is
$$\boxed{v_i = \sqrt{v_{i,x}^2 + v_{i,y}^2} = v \cdot \mathrm{cosec} (\theta)}$$
