Although the radiation-dominated (RD) era is long in comparison to the matter-dominated (MD) and $\Lambda$-dominated ($\Lambda$D) eras, it is nice to have an answer which can be adapted easily for any cosmological era. If we assume that the Universe is permeated by a perfect fluid we may use the equation of state
\begin{equation}
w = \frac{P}{\rho},
\end{equation}
where $P$ is the pressure and $\rho$ the energy density. The two Friedmann equations (or the conservation of the stress-energy tensor $\nabla_{\mu} T^{\mu \nu}=0$) give us
\begin{equation}
\dot{\rho} + 3H\rho(1+3w) = 0,
\label{F3}
\end{equation}
which can be solved for $\rho$ in terms of the scale factor $a(t)$ as
\begin{equation}
\rho = \rho_0 \Big(\frac{a}{a_0}\Big)^{-3(1+w)},
\end{equation}
where $a_0$ is the scale factor today (and we will set $a_0 = 1$ from now on) and $\rho_0$ is the total energy density of the Universe today. Plugging this back into the first Friedmann equation then gives $a(t)$ as
\begin{equation}
a(t) = At^{\frac{2}{3(1+w)}},
\end{equation}
where $A$ is a horrible constant which I have calculated to be $A = (8 \pi G/3)^{1/3(1+w)}$ and I have assumed that at some 'initial time' $t_i = 0$, the value of the scale factor was $a_i = 0$.
Now we consider the photon. As in my answer here, the photon follows a radial geodesic in FRW spacetime. In a flat Universe as you specified, we have
\begin{equation}
ds^2 = -dt^2 + dr^2 = 0
\end{equation}
for a photon. We can define an origin $(0,0,0)$ and from there send a photon to any point at radial distance $r = x$. You specified $(x,0,0)$ but as the Universe is homogeneous and isotropic, any such point will yield the same answer. Using the photon line element, this is
\begin{equation}
\int^x_0 dr = \int^{T_2}_{T_1} \frac{dt}{a(t)}
\end{equation}
Now we use the expression for $a(t)$ in terms of $t$ and $w$ which we derived earlier, evaluate the integrals in terms of $x$ and $t$, and set $T_1 = 0$ so we can see what is going on, arriving at
\begin{equation}
x = \frac{3(1+w)}{A(3w+1)}T_2^{1-\frac{2}{3(1+w)}},
\end{equation}
with the previously-stated value of $A$.
RD era ($w = 1/3$):
Substituting this value of $w$ we find
\begin{equation}
x \propto \sqrt{T_2} \implies T_2 \propto x^2,
\end{equation}
so given an infinite amount of time, we can send a signal out to infinity.
MD era ($w = 0$):
Substituting this value of $w$ we find
\begin{equation}
x \propto T_2^{1/3} \implies T_2 \propto x^3,
\end{equation}
so again we can send a signal to infinity but it will take $O(x)$ longer than in the RD era. This is because the Universe is expanding faster, as $a(t) \propto t^{2/3}$ rather than $a(t) \propto t^{1/2}$ in the RD era.
$\Lambda$D era ($w \rightarrow -1$):
For this case it is better to write $x$ in terms of $a$, in which case we obtain
\begin{equation}
x \propto a^{-1}.
\end{equation}
This means that as the Universe grows in the dark-energy regime, its expansion accelerates such that the region over which we can communicate shrinks! If we send out a photon now, the maximum distance it can reach in an infinite time will be greater than if we send out a photon tomorrow. This is the phenomenon of the shrinking Hubble sphere and it means that if the Universe continues to be dominated by dark energy as it is now, then the observable Universe will shrink until we can see only the closest astronomical objects.
