Here : Weight of block $w=2\, kN$,
Distance $d=10\, m$
Angle of inclination on the plane $\alpha=15^{\circ}$
The block will be pulled up on a smooth plane
Hence, force of resistance due to inclination
$F =w \sin \alpha=2 \times 10^{3} \sin 15^{\circ}$
$=2 \times 10^{3} \times 02588$
$=0.5176\, kN$
Now work done,
$W=F d =0.5176 \times 10^{3} \times 10$
$=5.17\, kJ$