Another alternative, that gives the same resuts, is to use parabolic cylindrical coordinates. In these coordinates, say $\tau, \sigma$ and $z$, we have the following relations to standard Cartesian coordinates:
\begin{eqnarray}
x &=& \sigma \, \tau \, ,
\\
y &=& \frac{1}{2} \left(\tau^2 - \sigma^2\right) \, ,
\\
z &=& z \, ,
\end{eqnarray}
with $\tau \in (-\infty,\infty), \, \sigma \in [0,\infty)$ and $z \in (-\infty,\infty)$. Fixing $\sigma = \sigma_0 \neq 0$ and $z = 0$, one finds curves given by
\begin{equation}
y = \frac{x^2}{2 \sigma_0^2} - \frac{\sigma_0^2}{2} \, ,
\end{equation}
which are parabolas with their foci in the origin. In terms of the paramenters of the problem, $d = \frac{\sigma^2_0}{2}$ or $\sigma_0 = \sqrt{2d}$. Notice that both $\sigma$ and $\tau$ have units of Length$^{1/2}$. The corresponding scale factors are
\begin{equation}
h_\sigma = h_\tau = \sqrt{\tau^2 + \sigma^2} \, \, \, \mathrm{and} \, \, \, h_z = 1 \, .
\end{equation}
Thanks to this info, we can write the corresponding charge density
\begin{equation}
\rho(\mathbf{x}') = \lambda \, \frac{\delta(\sigma' - \sqrt{2d})}{\sqrt{\tau'^2 + \sigma'^2}} \, \delta(z') \, .
\end{equation}
Now, we can compute the electric field at the origin with
\begin{equation}
\mathbf{E}(\mathbf{x} = 0) = - \frac{1}{4 \pi \epsilon_0} \int \mathrm{d}^3 x' \, \rho(\mathbf{x}')
\frac{\mathbf{x}'}{|\mathbf{x'}|^3} \, ,
\end{equation}
and
\begin{equation}
\mathbf{x}' = \sigma' \tau' \, \mathbf{i} + \frac{1}{2} \left(\tau'^2 - \sigma'^2\right) \, \mathbf{j} + z' \, \mathbf{k} \, .
\end{equation}
Notice also that $\mathrm{d}^3 x' = (\tau'^2 + \sigma'^2) \, \mathrm{d} \tau' \, \mathrm{d} \sigma' \, \mathrm{d} z'$. Integration basically goes as explained in other answers, meaning that one can simply focus on the $\mathbf{j}$ component of $\mathbf{E}(\mathbf{x} = 0)$, as the others vanish.
EDIT BONUS:
As a bonus, you can use this info to go back to the magnetic field problem. For a uniform current $I$ along the parabola you can write the current density as
\begin{equation}
\mathbf{J}(\mathbf{x'}) = I \, \frac{\delta(\sigma' - \sqrt{2d})}{\sqrt{\tau'^2 + \sigma'^2}} \, \delta (z') \, \, \boldsymbol{\tau'}
\end{equation}
where we used the unit vector (see the wiki link at the beginning of the answer)
\begin{equation}
\boldsymbol{\tau'} = \frac{\sigma' \, \mathbf{i} \, + \, \tau' \, \mathbf{j}}{\sqrt{\tau'^2 + \sigma'^2}} \, .
\end{equation}
Then you can use this to compute
\begin{equation}
\mathbf{B}(\mathbf{x} = 0 ) = - \frac{\mu_0}{4 \pi} \int \mathrm{d}^3 x'\, \mathbf{J}(\mathbf{x'}) \times \frac{\mathbf{x'}}{|\mathbf{x'}|^3} \, .
\end{equation}