1. Set up the rod and charge density

A uniformly charged thin rod has:

• length $L$
• total charge $Q$
• linear charge density

$\lambda = \frac{Q}{L}$

Let the rod lie along the $x$-axis from $x=-\frac{L}{2}$ to $x=\frac{L}{2}$, with its centre at the origin.

Point $P$ lies on the perpendicular bisector at distance

$a=\frac{\sqrt{3}}{2}L$

So the coordinates of $P$ are $(0,a)$.

2. Consider a small charge element

Take an element of length $dx$ at position $x$ on the rod. Then

$dq = \lambda \, dx$

Distance from this element to point $P$ is

$r = \sqrt{x^2 + a^2}$

The field due to $dq$ is

$dE = \frac{1}{4\pi\varepsilon_0}\frac{dq}{r^2}$

Because of symmetry, horizontal components cancel. Only the vertical components add.

Vertical component:

$dE_y = dE \cos\theta = \frac{1}{4\pi\varepsilon_0}\frac{dq}{r^2}\cdot \frac{a}{r}$

since

$\cos\theta = \frac{a}{r}$

Thus,

$dE_y = \frac{1}{4\pi\varepsilon_0}\frac{a\,dq}{(x^2+a^2)^{3/2}}$

Substitute $dq=\lambda dx$:

$dE_y = \frac{1}{4\pi\varepsilon_0}\frac{a\lambda\,dx}{(x^2+a^2)^{3/2}}$

3. Integrate over the rod

$E = \frac{1}{4\pi\varepsilon_0} a\lambda \int_{-L/2}^{L/2} \frac{dx}{(x^2+a^2)^{3/2}}$

Using symmetry,

$E = \frac{1}{4\pi\varepsilon_0} \cdot 2a\lambda \int_0^{L/2} \frac{dx}{(x^2+a^2)^{3/2}}$

Now use the standard integral:

$\int \frac{dx}{(x^2+a^2)^{3/2}} = \frac{x}{a^2\sqrt{x^2+a^2}}$

So,

$E = \frac{1}{4\pi\varepsilon_0} \cdot 2a\lambda \left[\frac{x}{a^2\sqrt{x^2+a^2}}\right]_0^{L/2}$

$E = \frac{1}{4\pi\varepsilon_0} \cdot 2a\lambda \cdot \frac{L/2}{a^2\sqrt{(L/2)^2+a^2}}$

$E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{\lambda L}{a\sqrt{(L/2)^2+a^2}}$

Since $\lambda L = Q$,

$E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{a\sqrt{(L/2)^2+a^2}}$

4. Substitute $a=\frac{\sqrt{3}}{2}L$

First,

$a^2 = \frac{3}{4}L^2$

Then,

$\left(\frac{L}{2}\right)^2 + a^2 = \frac{L^2}{4} + \frac{3L^2}{4} = L^2$

Hence,

$\sqrt{\left(\frac{L}{2}\right)^2 + a^2} = L$

So,

$E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{aL}$

Now put $a=\frac{\sqrt{3}}{2}L$:

$E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{\left(\frac{\sqrt{3}}{2}L\right)L}$

$E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2Q}{\sqrt{3}L^2}$

$E = \frac{Q}{2\sqrt{3}\pi\varepsilon_0 L^2}$

5. Match with options

This is exactly:

$\boxed{\frac{Q}{2\sqrt{3}\pi\varepsilon_0 L^2}}$

So the correct option is:

$\boxed{\text{C}}$