The interaction energy of London forces between two particles is proportional to $$r^x$$, where r is the distance between the particles. The value of x is:

London dispersion forces are weak intermolecular attractions that arise from momentary dipoles produced due to the movement of electrons in atoms or non-polar molecules.

The potential (interaction) energy $$E$$ of London forces varies with the separation $$r$$ between two particles according to the relation
$$$E \;\propto\; \dfrac{1}{r^{6}}$$$

Writing the proportionality in the form $$E \;\propto\; r^{\,x}$$ and comparing with $$$E \;\propto\; r^{-6}$$$, we obtain
$$x = -6$$

Hence, the exponent $$x$$ is $$-6$$, which corresponds to Option B.