Let the initial weight of the berries be denoted as \( W = 200 \) kg. We know that the berries are \( 98% \) water by weight. Therefore, the amount of water in the berries can be calculated as follows:
\[ \text{Weight of water} = 0.98 \times 200 = 196 \text{ kg} \]
The total weight of solids (non-water content) in the berries is:
\[ \text{Weight of solids} = 200 - 196 = 4 \text{ kg} \]
After two days in the sun, the berries are now \( 90% \) water. Let's denote the new total weight of the berries after two days as \( W' \). The weight of the water in this new situation can be expressed as:
\[ \text{Weight of water after two days} = 0.90 \times W' \]
The weight of the solids remains unchanged at \( 4 \) kg. Thus, we can express the total weight in terms of the water and solid component as follows:
\[ W' = \text{Weight of water} + \text{Weight of solids} = 0.90 W' + 4 \]
Now we can rearrange and solve for \( W' \):
\[ W' - 0.90 W' = 4 \] \[ 0.10 W' = 4 \] \[ W' = \frac{4}{0.10} = 40 \text{ kg} \]
Thus, the total weight of the berries after two days is
\[ \boxed{40} \] kg.