Answer
$\theta=39^o$
Work Step by Step
In an unbanked curve, static friction $f_s$ provides the centripetal force $F_c$ that keeps a vehicle from not skidding off. $F_c^{max}$ would, therefore, correspond to $f_s^{max}$. In other words, $$f_s^{max}=F_c^{max}$$ $$\mu_sF_N=\frac{mv_0^2}{r}$$
Since there is no vertical acceleration, $F_N=mg$ $$\mu_smg=\frac{mv_0^2}{r}$$ $$v_0^2=\mu_sgr (1)$$
In a banked curve, however, the horizontal component of the normal force $F_N\sin\theta$ provides the required $F_c$. An equation that represents this case is $$\tan\theta=\frac{v_0^2}{gr}$$ $$v_0^2=\tan\theta gr (2)$$
From (1) and (2), we have $$\tan\theta=\mu_s=0.81$$ $$\theta=39^o$$
