That's a long wire!

Let us consider the current to be moving in a clockwise nature. Let the coordinates of a general point on the parabola be defined by the following position vector

$\vec{r}_p = \frac{t^2}{4} \hat{i} + t \hat{j} + 0\hat{k}$

The position vector of the focus of the parabola is:

$\vec{r}_f = 1 \hat{i} + 0 \hat{j} + 0\hat{k}$

$\vec{r} = \vec{r}_p - \vec{r}_f$

From here, a vector directed tangentially to the parabola (directed along the current) can be computed as such:

$d\vec{L} = \left(\frac{t}{2} \hat{i} + 1 \hat{j} + 0\hat{k}\right)dt$

Now, Applying Biot-Savart Law:

$d\vec{B} = dB_x \hat{i} + dB_y \hat{j} + dB_z \hat{k} = \frac{\mu_oI}{4\pi}\left(\frac{d\vec{L}\times \hat{r}}{\mid \vec{r} \mid^2}\right)$

Substituting all expressions and simplifying leads to:

$dB_x = 0$ $dB_y = 0$ $dB_z = \frac{16}{(t^2+4)^2}dt$

From here:

$B_z = \int_{-\infty}^{\infty}\frac{16}{(t^2+4)^2}dt$

This integral has a closed form solution and can be solved by taking $t = 2 \tan(\theta)$ as $\theta$ varies from $-\pi/2$ to $\pi/2$ .

Therefore: $\boxed{B_z = \pi}$