Q:

How to prove the following relation $H=\frac{E\times v}{c}$

Using CGS units, how can we prove the following relation $$H=\frac{E\times v}{c}$$ where H is magnetic field, E is electric field and v is velocity

A:

What you have here is basically the B-field seen in the frame of a charge with velocity $\mathbf{v}$, moving in an electric field that is $\mathbf{E} \perp \mathbf{v}.$
You can derive this expression by considering the relativistic field transformations of $\mathbf{E}$ and $\mathbf{B}$ in a moving frame, I'll only show you the most important steps, for a complete walkthrough see chapter 26 of Feynman Lectures Vol. 2. Starting with the magnetic field $\mathbf{B}$ expressed in terms of the vector potential $\mathbf{A}$,:  $\mathbf{B} = \mathbf{\nabla}\times \mathbf{A},$ with the four-vectors in component being(using the short notation for partials): 
\begin{align*}
\nabla_{\mu} &= (\partial_t,-\partial_x,-\partial_y,-\partial_z)\\
\mathbf{A}_{\mu} &= (\phi,A_x,A_y,A_z) \text{ where } \phi = A_t
\end{align*}
Now we can write the outer product results for $\mathbf{B},$ to simplify we use a short-hand term like $B_y=F_{xz}= \partial_z A_x - \partial_x A_z,$ which in a generalized form is:
$$
F_{\mu \nu} = \nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}
$$
Next we Lorentz transform the $F_{\mu \nu}$ tensors, e.g. for $F'_{xy}$ we have: 
$$
F'_{xy} = \frac{F_{xy}-vF_{ty}}{\sqrt{1-v^2/c^2}}
$$
for shortness I will only write the magnetic field terms:
\begin{align*}
B'_x &= B_x \\
B'_y &= \frac{B_y+vE_z}{\sqrt{1-v^2/c^2}} \\
B'_z &= \frac{B_z-vE_y}{\sqrt{1-v^2/c^2}}
\end{align*}
With these transformations you can now calculate the B-field in a moving frame (with speed $v$ here). In order to get closer to your expression, we still have to express these in terms of vector products, for this we assume the velocity vector is directed in the positive $x$-direction, so you can e.g. rewrite $B_y +vE_z$ as the y-component of $(\mathbf{B}-\mathbf{v}\times \mathbf{E})_y,$ and so on so forth. Finally we just rewrite everything in terms of $\parallel$ (to $\mathbf{v}$) and $\perp$ components, with the field components along x-axis being the parallel ones and y,z the perpendicular ones. With the parallel terms being same in either frames, we have for the $\perp$ term (with $\gamma$ the Lorentz factor):
$$
B'_{\perp} = \gamma\left(\mathbf{B}-\frac{\mathbf{v}\times \mathbf{E}}{c^2}\right)_{\perp}
$$ 
From here, if you consider the special case of having no B-field in the rest frame, i.e. $\mathbf{B}=\mathbf{0},$ then the perpendicular components of $B'$ as seen in the moving charge's frame is: 
$$
B'_{\perp} = -\gamma\frac{\mathbf{v}\times \mathbf{E}}{c^2}
$$
Now absorb the Lorentz factor in E and rewrite it as a prime term, then remove the minus sign by using the anticommutative property of vector products and you're done.

