As you state, conformal time is defined as
$$
\eta(t) = \int_0^t\frac{\text{d}t'}{a(t')}.
$$
Using
$$
\dot{a} = \frac{\text{d}a}{\text{d}t},
$$
this can be written in the form
$$
\eta(a) = \int_0^a\frac{\text{d}a}{a\dot{a}} = \int_0^a\frac{\text{d}a}{a^2H(a)},
$$
with
$$
H(a) = \frac{\dot{a}}{a} = H_0\sqrt{\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}}.
$$
The scale factor $a$ is related to the redshift as
$$
1 + z = \frac{1}{a},
$$
so that
$$
\eta(z) = \frac{1}{H_0}\int_0^{1/(1+z)}\frac{\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a + \Omega_{K,0}\,a^2 + \Omega_{\Lambda,0}\,a^4}}.
$$
Basically, the conformal time is equal to the distance of the particle horizon, divided by $c$ (see this post for more info). $\eta_0$ refers to the current conformal age of the universe.
edit
I just checked the values that you posted with my own cosmology calculator. I get
$$
\eta_0 = 45.93\;\text{Gigayears}
$$
for the current conformal age of the universe, and
$$
\begin{align}
z_e &= 3234,& a_e &= 0.000309,& t_e &= 5.54\times 10^{-5}\;\text{Gy},&\eta_e &= 0.3804\;\text{Gy},\\
z_E &= 0.39,& a_E &= 0.719,& t_E &= 9.359\;\text{Gy},& \eta_E &= 41.08\;\text{Gy},\\
z_d &= 1089,& a_d &= 0000917,& t_d &= 0.00037\;\text{Gy},& \eta_d &= 0.911\;\text{Gy},
\end{align}
$$
so that
$$
\frac{\eta_e}{\eta_0} = 0.00828,\quad \frac{\eta_E}{\eta_0} = 0.894,\quad \frac{\eta_d}{\eta_0} = 0.0198.
$$
So my results are almost the same, but there's a small discrepancy. Apparently there's a small numerical error somewhere.
Best Answer
I am not sure this is what you want but I want to give it a try,
$$\eta=\int \frac {dt} {a}=\int\frac {da} {a\dot{a}}=\int\frac {da} {a^2H}$$ and we can write $$H(z)=H_0E(z)$$ $$E(z)=\sqrt{\Omega_{\Lambda}+\Omega_m(1+z)^3+\Omega_r(1+z)^4+\Omega_{\kappa}(1+z)^2}$$
so we have,
$$\eta=\int\frac {da} {a^2H_0E(z)}$$
and $dz=-da/a^2$ so we can write,
$$\eta=-H_0^{-1}\int\frac {dz} {E(z)}$$
And by taking initial coniditon as $z=\infty$, and due to the minus sign the integral becomes,
$$\eta=H_0^{-1}\int_z^{\infty}\frac {dz} {E(z)}$$
To find the current($t_0$) conformal time, we can use the above equation, for $z=0$
$$\eta=H_0^{-1}\int_{z=0}^{\infty}\frac {dz} {E(z)}$$
$$\eta=H_0^{-1}\int_{0}^{\infty}\frac {dz} {\sqrt{\Omega_{\Lambda}+\Omega_m(1+z)^3+\Omega_r(1+z)^4+\Omega_{\kappa}(1+z)^2}}$$
For the current values of $\Omega_{\Lambda}=0.69$, $\Omega_m=0.31$,$\Omega_{\kappa}= \Omega_r=0$
we have,
$$\eta=H_0^{-1}\int_{0}^{\infty}\frac {dz} {\sqrt{\Omega_{\Lambda}+\Omega_m(1+z)^3}}$$
$$\eta=H_0^{-1}\int_{0}^{\infty}\frac {dz} {\sqrt{0.69+0.31(1+z)^3}}$$
If we take $H_0=70km/s/Mpc$ then $1/H_0=1/(70\times 3.2408\,10^{-20})=4.4133353\,10^{17}s$
And the integral gives, $$\int_{0}^{\infty}\frac {dz} {\sqrt{0.69+0.31(1+z)^3}}=3.266054427285631$$
so $$\eta(t_0)=3.266054427285631\times 4.4133353\,10^{17}s=1.4414193\,10^{18}=45.70 \,\text {Gigayear}$$
To calculate the integral you can use, this site
I write the integral in terms of $z$ but, its also possible to write the equation in terms of $a(t)$ (the begining part of the derivation). But $z$ is the observable value so I prefer to write in that form.
For a given $t$ you can turn easily $a(t)$ to $z$.