1. Given charge distribution

The rod lies along the $x$-axis from $x=0$ to $x=l$, with linear charge density

$\lambda(x)=\rho(x)=\rho_0\frac{x}{l}.$

It rotates about an axis through $x=0$ and perpendicular to the rod with frequency $n$ rotations per second.

So angular speed is

$\omega = 2\pi n.$

2. Take a small element of the rod

Consider a small element of length $dx$ at distance $x$ from the axis.

Its charge is

$dq = \lambda(x)\,dx = \rho_0\frac{x}{l}\,dx.$

As the rod rotates, this small charge moves in a circle of radius $x$.

3. Current due to the rotating charge element

If a charge $dq$ completes $n$ revolutions per second, the equivalent current is

$dI = n\,dq.$

Thus,

$dI = n\rho_0\frac{x}{l}\,dx.$

4. Magnetic moment of this current loop

A current loop has magnetic moment

$d\mu = dI \times \text{area}.$

Here the area of the circular path is

$A = \pi x^2.$

So,

$d\mu = dI\,\pi x^2 = \left(n\rho_0\frac{x}{l}\,dx\right)\pi x^2.$

Therefore,

$d\mu = \pi n \rho_0 \frac{x^3}{l}\,dx.$

5. Integrate over the rod

$\mu = \int_0^l d\mu = \pi n \rho_0 \frac{1}{l} \int_0^l x^3\,dx.$

Now,

$\int_0^l x^3\,dx = \frac{l^4}{4}.$

Hence,

$\mu = \pi n \rho_0 \frac{1}{l}\cdot \frac{l^4}{4} = \frac{\pi}{4} n \rho_0 l^3.$

6. Final answer

Thus the time-averaged magnetic moment is

$\boxed{\mu = \frac{\pi}{4} n \rho_0 l^3}.$

This matches Option B.

Note: The magnetic moment is constant in magnitude and directed along the axis of rotation, so its time average is the same value.