Correct Option: (c)

The key to solving this problem efficiently is choosing the correct frame of reference. The river flow is uniform, and the wooden block floats with the river. Therefore, in the frame of reference of the water (river):

• The water is stationary.
• The wooden block is stationary (velocity = 0).
• The boat moves with a constant speed $v_{rel}$ relative to the water.

The energy expended by the rower is primarily used to do work against the viscous drag force of the water. Since the boat moves at a constant speed relative to the water, the viscous force $F_{viscous}$ is constant in magnitude.

The work done is given by:

$$ E = \int \vec{F}_{viscous} \cdot d\vec{r}_{rel} = F_{viscous} \times (\text{Distance moved relative to water}) $$

Let’s analyze the displacement relative to the water in each case:

1. Downstream: The boat moves 20 m ahead of the block. Since the block is our origin in this frame, displacement $d_1 = 20 \text{ m}$.
2. Upstream: The boat moves 20 m away from the block. Displacement magnitude $d_2 = 20 \text{ m}$.
3. Sideways: The boat moves with the stream (longitudinally stationary relative to the block) and shifts 20 m towards the bank. Displacement magnitude $d_3 = 20 \text{ m}$.

Since the distance covered relative to the water is $20 \text{ m}$ in all three cases, the work done against the viscous force is identical.

$$ E_1 = E_2 = E_3 $$