Inspired from Steven Sir!

Let the position vector of a point on the vertical wire be:

$\vec{r}_p = y \hat{j}$

And that on the upper half of the right branch of the hyperbola be:

$\vec{r}_h = x \hat{i} + \sqrt{x^2-4} \hat{j}$

The vector joining these two points directed away from the wire is:

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

We consider a line element on the wire. The length of the element near the given position vector has charge $\lambda dy$ . We also consider an arc length element on the hyperbola. The charge on that element near the given position vector is $\sigma ds$ where $ds$ is the arc length magnitude which is:

$ds = dx\sqrt{1 + \left(\frac{dy}{dx}\right)^2}$

For the upper half of the right lobe:

$ds = dx\sqrt{\frac{2x^2-4}{x^2-4}}$

Applying Columb's law to find electrostatic force at wire element due to the hyperbolic element:

$d\vec{F} = \frac{K \left(\lambda dy\right)\left(\sigma ds\right)\vec{r}}{\lvert \vec{r} \rvert^3}$

$K = \frac{1}{4\pi \epsilon_o}$

$d\vec{F} = dF_x\hat{i} + dF_y \hat{j}$

Substituting all expressions and simplifying gives:

$dF_x = \left(\frac{K \lambda \sigma\sqrt{2}\,x\,\sqrt{\frac{2\,x^2-4}{2\,\left(x^2-4\right)}}}{{\left({\left(y-\sqrt{x^2-4}\right)}^2+x^2\right)}^{3/2}}\right)dx \ dy$

Similarly an expression for $dF_y$ can be computed. Now, the total force on the wire due to the upper half of the right lobe of the hyperbola is:

$F_x = \int_{-2}^{2} \int_{2}^{\infty}\left(\frac{K \lambda \sigma\sqrt{2}\,x\,\sqrt{\frac{2\,x^2-4}{2\,\left(x^2-4\right)}}}{{\left({\left(y-\sqrt{x^2-4}\right)}^2+x^2\right)}^{3/2}}\right)dx \ dy$

Therefore, by symmetry, the force due to the entire right lobe of the hyperbola will be:

$F_R = 2F_x$

Again, by symmetry, the force due to both lobes along the x-direction will be:

$F_T = 2F_R = 4F_x$

Therefore, the acceleration $A$ is:

$A = \frac{F_T}{M}$

All forces along Y will evaluate to zero. One can check this by even evaluating the integrals. This step is left out here.

Having obtained $F_T$ , the answer comes out to be:

$\boxed{20\frac{A}{K} \approx 410.373}$