Answer
$v_{max} = \sqrt (Rg+ tan( \theta+ tan^{-1} μ_{s}))$
$ = \sqrt \frac{Rg tan( \theta+ μ_{s})}{1- μ_{s} tan\theta}$
Work Step by Step
To be on the verge of sliding out means that the force of static friction is acting with maximum magnitude. We first consider the vector sum of force $F$ equals the static friction force and the normal force. Due to the facts that they are perpendicular, $\theta $ is at an angle measured from the vertical axis
$ϕ _{s} =θ+ \theta_{s} $
$tanθ = μ_{s}$ and θ is the bank angle.
Now, the vector sum of force $F$ and the vertically downward pull (mg) of gravity must be equal to the (horizontal) centripetal force which leads to a surprisingly simple relationship
$tan ϕ = \frac{\frac{mv^{2}}{R}}{mg}$
$= \frac{v^{2}}{Rg}$
For maximum speed :
$v_{max} = \sqrt (Rg+ tan( \theta+ tan^{-1} μ_{s}))$
$ = \sqrt \frac{Rg tan( \theta+ μ_{s})}{1- μ_{s} tan\theta}$
