Resolve the forces perpendicular and parallel to the surface of the incline. They must separately be balanced.
Perpendicular
N = 1.85*9.8*cos(67.7°) + F*sin(67.7°) ...(1) N is normal force, gravitational accn = 9.8m/s2
Parallel
0.256*N + F*cos(67.7°) = 1.85*9.8*sin(67.7°) ...(2)
Use (1) in (2)
0.256*(1.85*9.8*cos(67.7°) + F*sin(67.7°)) + F*cos(67.7°) = 1.85*9.8*sin(67.7°)
F*(0.256*sin(67.7°) + cos(67.7°)) = 1.85*9.8*(sin(67.7°) - 0.256*cos(67.7°) )
F = 1.85*9.8*(sin(67.7°) - 0.256*cos(67.7°) )/(0.256*sin(67.7°) + cos(67.7°))
$${\mathtt{F}} = {\frac{{\mathtt{1.85}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{67.7}}^\circ\right)}{\mathtt{\,-\,}}{\mathtt{0.256}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{67.7}}^\circ\right)}\right)}{\left({\mathtt{0.256}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{67.7}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{67.7}}^\circ\right)}\right)}} \Rightarrow {\mathtt{F}} = {\mathtt{24.359\: \!321\: \!813\: \!047\: \!781\: \!3}}$$
F ≈ 24.4 Newtons
