Answer
In this situation, we take $\vec{f}_{s}$ to point uphill and to be equal to its maximum value, in which case $f_{s, \max }=\mu_{s} F_{N}$ applies, where $\mu_{s}=0.25 .$ Applying second law for Newton's to the block of mass $m=W / g=8.2 \mathrm{kg},$ in the $x$ and $y$ directions, produces
$$
F_{\min 1}-m g \sin \theta+f_{s, \max }=m a=0$$$$
F_{N}-m g \cos \theta=0
$$
$\left.\text { which (with } \theta=20^{\circ}\right)$ leads to
$$
F_{\min 1}-m g\left(\sin \theta+\mu_{s} \cos \theta\right)=8.6 \mathrm{N}
$$
Work Step by Step
In this situation, we take $\vec{f}_{s}$ to point uphill and to be equal to its maximum value, in which case $f_{s, \max }=\mu_{s} F_{N}$ applies, where $\mu_{s}=0.25 .$ Applying second law for Newton's to the block of mass $m=W / g=8.2 \mathrm{kg},$ in the $x$ and $y$ directions, produces
$$
F_{\min 1}-m g \sin \theta+f_{s, \max }=m a=0$$$$
F_{N}-m g \cos \theta=0
$$
$\left.\text { which (with } \theta=20^{\circ}\right)$ leads to
$$
F_{\min 1}-m g\left(\sin \theta+\mu_{s} \cos \theta\right)=8.6 \mathrm{N}
$$
