Answer
$8.6^{\circ}$; the mass of the dice drops out of the equations.
Work Step by Step
We know that
$\theta=tan^{-1}(\frac{F_x}{F_y})$
$\implies \theta=tan^{-1}(\frac{\frac{mv^2}{r}}{mg})$
$\implies \theta=tan^{-1}(\frac{v^2}{gr})$
We plug in the known values to obtain:
$\theta=tan^{-1}(\frac{(27mi/h\times 0.447\frac{m/s}{mi/h})^2}{(9.81)(98)})$
$\theta=tan^{-1}(0.1515)$
$\theta=8.6^{\circ}$
We can see that the mass of the dice is cancelled out. Hence, it is unnecessary to give the mass of the dice.
