Radiation force is totally negligible at the surface of the Sun - about one hundred thousandth the force due to gravity.
The radiation force does grow as you move into the solar interior, but only as the inverse square of the radius considered, but will not increase much inside the radius at which the solar luminosity begins to be produced (the core). As a result, the radiation force will still be about a ten thousandth of gravity or less inside the Sun.
In order to increase the influence of radiation you need to increase the surface area exposed to that radiation whilst keeping the mass (and hence gravitational force) of the object constant.
(Note added: I took you question to be literally about radiation pressure due to light. If you replace "radiation pressure" with just "pressure", then indeed the point where you would "float" is where the density is about that of the human body, halfway into the solar interior. This is because at that point the pressure gradient balances the gravitational force for material of that density. However, the gas pressure at this depth is about a billion atmospheres, so you would be crushed.)
Details:
The flux of radiation (power per unit area) from the surface of a blackbody is $\sigma T^4$, where $T$ is the temperature and $\sigma$ is the Stefan-Boltzmann constant.
The force due to radiation is the radiation flux integrated over the surface area $A$ that an object presents to the radiation, and then divided by the speed of light, $c$.
The force due to gravity is just $mg$, where $m$ is the mass of the body and $g$ is the local gravity.
The ratio of the two is given algebraically by
$$ \frac{F_r}{F_g} = \frac{ A \sigma T^4}{mg c} $$
and for a spherical object of mass $M$ inside radius $R$, then the surface gravity is $g= GM/R^2$, so
$$ \frac{F_r}{F_g} = \frac{ A \sigma T^4 R^2}{GMm c}\ . $$
At the surface of the Sun $T=5800$ K, $\sigma = 5.7\times 10^{-8}$ in SI units and we could assume that a person presents an area $A \simeq 1$ m$^2$ and has a mass of 80 kg. The mass of the Sun $M=2.0\times 10^{30}$ kg and the radius $R = 7.0\times 10^{8}$ m. Putting these numbers in, we find
$$\frac{F_r}{F_g} \simeq 10^{-5} $$
i.e. That the radiation force is a very small fraction of the gravitational force and could probably be ignored for most purposes.
You might then have thought that because the temperature in the interior of the Sun is much hotter (approaching 15 million K at the centre), that the radiation force/radiation pressure would become much more important compared to gravity. That is not the case. The radiation flux coming from the centre is just the luminosity of the Sun divided by the surface area of a sphere at the solar interior radius considered (note that this is true when the energy is carried solely by radiation and takes full account of the fact that radiation is hitting you from all sides). That is because the solar luminosity is all generated by fusion reactions in the core. In fact if you get very close to the centre then only a fraction of the luminosity should be considered becase some of the energy generation would be "above" you. The solar gravity would also change because you need only consider the mass interior to your radius (which goes down) and because the radius you are at is getting smaller. The ratio of radiation force to gravity becomes
$$\frac{F_r}{F_g} = \frac{A L R^2}{4\pi GM(<R)mc R^2}\ , $$
where $M(<R)$ is now the mass inside radius $R$. You can see that the $R^2$ terms cancel and so we see that this ratio will grow as you move into the Sun, since $M(<R)$ would get smaller, at least until you get to the core at about $R= 0.2R_\odot$, when $L$ would start to decrease too.
Now, you should be able to convince yourself that if you use $R=7\times 10^8$ m, $L= 3.8\times 10^{26}$ W and $M(<R)-2\times 10^{30}$ kg, appropriate for the surface, then you get the same answer as before. The mass of the Sun is however quite centrally concentrated so even if you move close to the solar core with $R \simeq 10^7$ m, you are only going to increase this fraction by a factor ofa few and thus the force due to radiation is still a small fraction of that due to gravity. An exact answer would need a detailed model of the run of $M(<R)$ with $R$ and of $L(<R)$ with $R$.
A further complication is that in the outer 20% of the Sun, the energy is mostly carried by convection rather than radiation (see https://astronomy.stackexchange.com/a/43196/2531). In this convective zone, the radiative forces will be even lower than estimated above by factors of a few.
