Consider a positive point charge moving in positive $z$ direction at a constant velocity, at that instant when it's at the origin. The magnetic-field lines generated in plane $z=a$ are closed loops, further we observe circular symmetry in the problem about $z$-axis, henceforth the loops are circles.
A image to develop a idea of how we would be going to proceed:
Let's apply Ampere's displacement law to calculate magnetic field. Intuitively we can think, as the particle moves towards the plane $z=a$, the electric flux passing through a surface bounded by any circle centered on the z-asix in that plane will increase.
So, starting with Ampere's Law we have:
$$\oint \vec{B}.d\vec{I} = \mu_o \epsilon_o \frac{d\phi}{dt}$$
We integrate the left hand-side around a circle of radius $b$ on $z$-axis in plane $z=a$, and this is our Amperian Loop.
so,
$$ \oint \vec{B}.d\vec{I} = 2 \pi bB$$
To calculate electric flux $\phi$ enclosed by the circle of radius $b$, we select a spherical surface, of surface area $A$, which is bounded by the circle, is symmetric about z-axis, and has a radius $r$. So it follows from the understanding, we have:
$$ \phi = \int \vec{E}.d\vec{A} = EA$$, where $$E= \frac{q}{4 \pi \epsilon_o r^{2}}$$
The area $A$ can be obtained using spherical, polar coordinates:
$$ A= r^{2} \int_0^2\pi d\phi \int_0^\theta \sin x dx = 2 \pi r^2(1-\cos \theta)$$
$$\cos \theta= \frac{z}{\sqrt{z^2+x^2}}$$
Using/combining the above three equations we can have:
$$\phi = \frac{q}{2 \epsilon_o}(1-\cos \theta)$$
Differentiating with time gives,
$$\frac{d\phi}{dt} = -\frac{q}{2 \epsilon_o} \frac{d \cos \theta}{dt}$$ where, $$\frac{d \cos \theta}{dt} = \frac{d \cos\theta}{dz} \frac{dz}{dt}$$
In the above equation, $z$ is the distance between the particle, which is moving in the positive $z$ direction, and the Amperian Loop, which is fixed in $z=a$ plane. Since $z$ is decreasing with time,
$$\frac{dz}{dt} = -v$$ where v is the speed of the particle. Now,
$$\frac{d \cos \theta}{dz}=\frac{y^2}{r^3}$$, where $r=\sqrt{z^2+y^2}$
Combining the first two and last three equations, we arrive at our awaited result as,
$$B= \frac{\mu_o}{4\pi} \frac{qv \sin \theta}{r^2}$$, which can be vectorially rewritten as, $$\vec{B} = \frac{\mu_o}{4\pi} \frac{q \vec{v} \times \vec{r}}{r^3}$$
Hence, we are done! :)
