Answer
$\approx 124\;\rm km/h$
Work Step by Step
We need to find the maximum speed the truck can take the curve without sliding which means we need to apply the maximum static friction force between the tires and the road.
But first, we need to draw a force diagram of the truck, as shown below.
As seen below, the radial acceleration is on the chosen direction of positive $x$-axis.
Thus,
$$\sum F_x=f_s\cos\theta+F_n\sin\theta=ma_x=ma_r$$
and we know that the radial acceleration is given by $v^2/R$.
$$ f_s\cos\theta+F_n\sin\theta=m\dfrac{v^2}{R}$$
We also know that the maximum static friction is given by $f_s=\mu_sF_n$
$$ \mu_sF_n\cos\theta+F_n\sin\theta=m\dfrac{v^2}{R}$$
Thus,
$$F_n\left(\mu_s \cos\theta+ \sin\theta\right)=m\dfrac{v^2}{R}\tag 1$$
$$\sum F_y=F_n\cos\theta-f_s\sin\theta-mg=ma_y=m(0)=0$$
Thus,
$$F_n\cos\theta-f_s\sin\theta=mg$$
$$m=\dfrac{F_n\cos\theta-f_s\sin\theta}{g}$$
$$m=\dfrac{F_n\cos\theta-\mu_sF_n\sin\theta}{g}$$
$$m=\dfrac{F_n\left(\cos\theta-\mu_s \sin\theta\right)}{g}$$
Plugging into (1);
$$ \color{red}{\bf\not}F_n\left(\mu_s \cos\theta+ \sin\theta\right)=\dfrac{ \color{red}{\bf\not}F_n\left(\cos\theta-\mu_s \sin\theta\right)}{g}\;\dfrac{v^2}{R} $$
Thus,
$$v^2=\dfrac{gR\left(\mu_s \cos\theta+ \sin\theta\right)}{\left(\cos\theta-\mu_s \sin\theta\right)}$$
$$v =\sqrt{\dfrac{gR\left(\mu_s \cos\theta+ \sin\theta\right)}{\left(\cos\theta-\mu_s \sin\theta\right)}}$$
Plugging the known;
$$v =\sqrt{\dfrac{9.8\times 70\left(\cos15^\circ + \sin15^\circ \right)}{\left(\cos15^\circ - \sin15^\circ \right)}}$$
$$v_{max}=\color{red}{\bf 34.5}\;\rm m/s\approx \color{red}{\bf124}\;\rm km/h$$
