A small block of mass 200 g is situated at the top of a...

Mass of the block, $m = 200 \text{ g} = 0.2 \text{ kg}$

Height of the incline, $h = 3.2 \text{ m}$

Length of the incline, $L = 10 \text{ m}$

Acceleration due to gravity, $g = 10 \text{ m/s}^2$

Since the incline is frictionless, the work done to slide the block up the incline slowly is equal to the change in its gravitational potential energy.

Alternatively, the force required to slide the block up the incline is $F = mg \sin \theta$, where $\sin \theta = \dfrac{h}{L}$.

Work done, $W = F \times L = mg \sin \theta \times L = mg \left( \dfrac{h}{L} \right) L = mgh$

$W = 0.2 \times 10 \times 3.2 = 6.4 \text{ J}$