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.
Best Answer
\begin{align*} ma &= -\mu mg \\ a &= -\mu g \\ \frac{v-u}{t} &= -\mu g \\ \frac{0-u}{3} &= -0.8 \times 9.8 \\ u &= 23.5 \end{align*}