How can I compute the minimum possible bank angle for a turn with a given radius? I know that I can find an upper bound for the bank angle for a plane, if I know the maximum sustained load factor of that plane because the roll angle and the load factor $n_z$ in a turn are directly related:
$$tan (\varphi) = \sqrt{n_z^2 - 1} \;\; \text{or} \;\; \varphi = arctan\left(\sqrt{n_z^2 - 1}\right)$$
And also I known that the bank angle would not be more than 30° for civilian and 75° and more for military aircraft.

But my question is :

How can one get the minimum possible bank angle an airplane can use to perform a turn with a given radius before reaching the stall speed?

Also I want to know if the stall speed is known, can we use the formula :

$$ W = V/R $$

to calculate the angular rate of turn (of course considering the level turn)?
Given a radius $R$ and a velocity $V$, the bank angle can be computed by: 

$\phi = \textrm{atan}\left(\frac{V^2}{R\cdot g }\right)$,

where $g$ is the acceleration by gravity. This shows that the lower the speed becomes, the lower the bank angle becomes. The lowest attainable speed is the stall speed, however we have to take into account the load factor during the turn. 

The load factor $n$ during the turn is:

$n=\frac{1}{\cos \phi}$

Due to the increased load factor, the stall speed increases in the turn:

$V_{stall,turn} = V_{stall} \cdot \sqrt n $

The minimum bank angle for a turn with radius R will occur when the speed is equal to the stall speed for that angle. 

Solving this mathematically for $\phi$ results in:

$\phi_{min} = \textrm{atan}\left(\frac{V_{stall}^2}{\sqrt{R^2g^2-V_{stall}^4}}\right)$

Note that is the theoretical approach; in practice you would not fly at the stall speed, especially not during turns. 

Units are all in SI, you have to apply conversion factors if you want to use aviation units.



Example:


$V_{stall}$ = 50 m/s (approx. 97 knots) 
$R$ = 400 m 
$\phi_{min}$ = ~39.6 degrees 
$V$ = ~57 m/s (approx. 111 knots)