For some equations, check out http://en.wikipedia.org/wiki/Banked_turn#Banked_turn_with_friction.
Of particular interest are the equations for minimum and maximum speed.
$$v_{max}=\sqrt{\frac{rg(\tan{\theta}+\mu_s)}{1-\mu_s\tan{\theta}}}$$
$$v_{min}=\sqrt{\frac{rg(\tan{\theta}-\mu_s)}{1+\mu_s\tan{\theta}}}$$
If you were to eliminate friction, you'd get
$$v_{frictionless}=\sqrt{rg\tan{\theta}}$$
You'll find that when there's no or nearly no friction, you have to maintain a very specific speed (similar to an orbit). When you increase the coefficient of friction, it increases the speed band in which you can go, allowing you to go both slower and faster without sliding down the slope or flying off of it respectively. So in terms of your minimum coefficient of friction, if changing $d$ gets you closer to $v_{frictionless}$, you don't need as much friction. If the change in $d$ gets you further from $v_{frictionless}$, you need more friction.
EDIT: To instead have the equations to find $\mu_s$
$$v_{max}^2(1-\mu_s\tan\theta)=rg\tan\theta+rg\mu_s$$
$$v_{max}^2-v_{max}^2\tan\theta\mu_s=rg\tan\theta+rg\mu_s$$
$$v_{max}^2-rg\tan\theta=v_{max}^2\tan\theta\mu_s+rg\mu_s$$
$$v_{max}^2-rg\tan\theta=(v_{max}^2\tan\theta+rg)\mu_s$$
$$\frac{v_{max}^2-rg\tan\theta}{v_{max}^2\tan\theta+rg}=\mu_s$$
Similarly
$$\frac{rg\tan\theta-v_{min}^2}{v_{min}^2\tan\theta+rg}=\mu_s$$
If your desired speed is higher than $v_{frictionless}$, use the version with $v_{max}$. Otherwise use the version with $v_{min}$. This will give you the minimum coefficient of friction needed to maintain that speed.
Of note is that
$$\frac{v_{max}}{v_{frictionless}}=\frac{v_{frictionless}}{v_{min}}$$
which gives you
$$v_{max}=\frac{v_{frictionless}^2}{v_{min}}$$
and
$$v_{min}=\frac{v_{frictionless}^2}{v_{max}}$$
So if you'd like to work with one of $v_{max}$ or $v_{min}$ plus $v_{frictionless}$, it can be done.
The problem of a banked turn with friction is worked out on Wikipedia, where the following equation for maximal velocity is derived:
$$
v= {\sqrt{rg\left(\sin \theta +\mu_s \cos \theta \right)\over \cos \theta -\mu_s \sin \theta }}
={\sqrt{rg\left(\tan\theta +\mu_s\right)\over 1 -\mu_s \tan\theta}}.
$$
This equation expresses the maximum velocity $v$ in terms of the angle of incline $\theta$, coefficient $\mu_s$ of static friction, and radius $r$ of curvature; $g\approx9.8\,{\rm m}/{\rm s}^2$.
Squaring both sides of the equation and isolating $\mu_s$ we find
$$
\mu_s = {v^2-rg\tan\theta\over v^2\tan\theta+rg}.
$$
Now it remains to substitute the given values:
$$
\theta=31^\circ, \quad
v=322\,{\rm km}/{\rm h}\approx 89.44\,{\rm m}/{\rm s}, \quad
r=304.8\,{\rm m}, \quad
g\approx9.8\,{\rm m}/{\rm s}^2.
$$
Substitution yields
$$
\mu_s \approx 0.796
$$
Best Answer
The centripetal motion is supplied by the force of friction $f = \mu N = \mu mg$. Setting equal $$ \frac{mv^2}{r} = \mu mg \to v = \sqrt{\mu g r} $$