There is no relationship of the form you have in mind.
The main point is that photons with the same wavelength do not have to be identical. For example, if an atom or nucleus emits a photon, we can further classify the photon according to its angular momentum and polarization.
In a semiclassical description, suppose that a classical sinusoidal plane wave packet has a certain given energy. We can't use that information to find the amplitude of the wave, because it depends on the dimensions of the wave train. We could have a low energy density over a large volume, or a high energy density over a small volume.
It is possible to cook up a semiclassical estimate for the greatest field that any photon of a given wavelength can have. Suppose that the photon is compressed to a volume $~\lambda^3$, which is on the order of the smallest volume a wave can have, for that wavelength. Then multiplying the energy density of the electric field by that volume and setting it equal to $hc/\lambda$ gives the estimate $|\textbf{E}|\sim \lambda^{-2}(khc)^{1/2}$, where $k$ is the Coulomb constant. (And in SI units, $|\textbf{B}|=|\textbf{E}|/c$.) For visible light, this is something like $10^5\ \text{V}/\text{m}$. But that should not be interpreted as "the" intensity of the field. E.g., for a photon of visible light from a laser, a realistic estimate of the volume is many orders of magnitude greater than $\lambda^3$, due to the long coherence length, so the field strength would be many orders of magnitude lower than this upper bound.
I tried a search and could not find it in a simple format. Like if the wavelength halves than the magnetic field falls off like $~\frac{1}{\lambda^2}$.
Yes, the estimate does fall off with that exponent -- but only the estimate for the maximum intensity that any photon can have. E.g., for a wave packet with fixed dimensions, the exponent would $-1/2$ rather then $-2$.
They use photons in laser traps and then make a beautfl picture of the electric field as all photons are added. So i was thinking how would the field of a single one be. I mean we can divide by the big number of photons, that are present.
For example, suppose we have a standing wave in a resonant microwave cavity, with volume $\sim\lambda^3$. Then the estimate above is of the right order of magnitude, and replacing the single-photon energy $hc/\lambda$ with $nhc/\lambda$ gives $|\textbf{E}|\sim \lambda^{-2}(nkhc)^{1/2}$. That is, the scaling is like $\sqrt{n}$, not $n$, because the photons are coherent.
