Answer
$28^{\circ}$
Work Step by Step
We can find the required angle as follows:
$tan\theta=\frac{v^2}{rg}$
$\theta=tan^{-1}\frac{v^2}{rg}$
We plug in the known values to obtain:
$\theta=tan^{-1}[\frac{(390Km/h\times 0.278\frac{m/s}{Km/h})^2}{(2300)(9.81)}]$
$\theta=28^{\circ}$