We can do a quick "back of the envelop calculation" to gauge the magnitude of this effect. For this I will make many simplifications, but it should give a good enough estimate.
Let's suppose that the Earth is a perfect mirror, (in reality it reflects roughly 30% of received light, so it would be better to assume it absorbs everything, but let's take the upper bound here). Let us also suppose that the Earth is a disc, so that every photon of the sun hits her perpendicularly.
According to Wikipedia, the flux emitted by the sun and received by Earth is $1362$ $W/m^2$, let's round that down to $1000 W/m^2$. Taking $6000$ $km$ for Earth's radius, the surface of our disc is roughly $A = \pi (6\cdot10^{6})^2m^2\approx 10^{14}m^2$. So the power radiated on Earth is $\approx 10^{17} W$.
How do we translate this information into a force exerted by the photons ? Each photon will have energy $E = h\nu$, where $\nu$ is it's frequency. If we suppose all photons have the same frequency, then there are $\frac{10^{11}}{h\nu}$ photons$/s$ hitting the Earth. Their momentum is simply $p = \frac{E}{c}$.
When they are reflected on the surface of the Earth, their momentum is flipped sign, so they undergo a change in momentum of $\delta p = 2\frac{E}{c}$. 187
We know by <PERSON>'s law that they must have been acted on by a force $F$ satisfying $F=\frac{\delta p}{\delta t} = \frac{2E}{c\delta t}$. But by <PERSON>'s third law, that means they acted on Earth by a force of the same magnitude !
We now have all the pieces. In a time $\delta t$, $N = \frac{10^{11}}{h\nu}\delta t$ photons will hit the Earth, each exerting a force $F = \frac{2E}{c\delta t}$ on it. The total force applied to the earth during that time is thus $F\times N = 10^{17}\frac{2E\not{\delta t}}{h\nu\not{\delta t}}=10^{17}\frac{2(\not{h\nu})}{c\not{h\nu}}=\frac{2}{c}10^{17} \approx 10^9 N$
As we can see, $\nu$ drops out conveniently so we don't have to assume anything about the frequency of the photon emitted by the sun. So there we have it, the force of the solar radiation is $10^9$ $N$, which is by no means a "weak" force, at least from a human's point of view. But let's compare it to something more comparable, as for example the force exerted by the Moon on the earth.
$ F_{moon} = G \frac{M_{moon}M_{Earth}}{r_{moon-earth}^2} \approx 10^{20} N$
So the force exerted by the Sun's photon are $0.000000001$% of the force exerted by the moon. As you can see, the effect is ridiculously small, and that is compared to the force exerted by the moon, which is in turn ridiculously small compared to the force exerted by the Sun on Earth.
In other words, I don't think we would be able to estimate the effects on Earth's orbit anytime soon ! (although it would be interesting to estimate what kind of difference in the orbit radius we are talking about, I will probably do the computation if I get some time !) 319