Answer
$15.3\;\rm km$
Work Step by Step
We know that Rayleigh’s criterion states that two objects separated by an angle $\alpha$ a are marginally resolvable when $\alpha=\theta_{\rm min}$ where $\theta_{\rm min}=1.22\lambda/D$ is the angular resolution of a lens of diameter $D$.
Thus,
$$\alpha=\dfrac{1.22\lambda}{D}$$
where $\lambda$ here is the wavelength inside your eye, so $\lambda=\lambda_{\rm air}/n_{\rm eye}$
$$\alpha=\dfrac{1.22\lambda_{\rm air}}{n_{\rm eye}D_{\rm pupil}}\tag1$$
For small angle approximation, $\tan\theta\approx\theta$, so
$$\alpha=\dfrac{d}{L}\tag 2$$
where $d$ is the separation distance between the two headlights of the car and $L$ is the distance between the headlights and your eye at which the two headlights are marginally resolved.
From (1) and (2),
$$\dfrac{d}{L}=\dfrac{1.22\lambda_{\rm air}}{n_{\rm eye}D_{\rm pupil}} $$
Solving for $L$,
$$L=\dfrac{n_{\rm eye}D_{\rm pupil} d} {1.22\lambda_{\rm air}}$$
Plugging the known;
$$L=\dfrac{(1.33)(7\times 10^{-3}) (1.20)} {1.22(600\times 10^{-9})}=\bf 1.53\times 10^4 \;\rm m$$
$$L=\color{red}{\bf 15.3}\;\rm km$$
