Answer
$$
F=3.0 \times 10^{7} \mathrm{N}
$$
Work Step by Step
Applying Newton's second law, we have
$$
F+m g \sin \theta-f_{s, \max } =m a=0 $$$$
F_{N}-m g \cos \theta=0
$$
Along with Eq. $6-1\left(f_{s, \max }=\mu_{s} F_{N}\right)$ we have enough information to solve for $F .$ With $\theta=24^{\circ}$ and $m=1.8 \times 10^{7} \mathrm{kg}$ , we find
$$
F=m g\left(\mu_{s} \cos \theta-\sin \theta\right)=3.0 \times 10^{7} \mathrm{N}
$$