Given:
Initial water content: \( w_1 = 0.15 \)
Degree of saturation: \( S = 67\% \)
Void ratio: \( e = 0.6 \)
Specific gravity of solids: \( G_s = 2.67 \)
Unit weight of water: \( \gamma_w = 9.81 \, \text{kN/m}^3 \)
Let water content after full saturation be \( w_2 \). The weight of water after full saturation can be calculated using the following steps.
Step 1: Water content at full saturation (\( w_2 \))
\[
w_2 = \frac{e}{G_s} = \frac{0.6}{2.67} = 0.2247
\]
Step 2: Change in weight of water (\( w_2 - w_1 \))
\[
w_2 - w_1 = \frac{\text{Weight of water}}{\text{Weight of solid}} = \frac{w}{w_s}
\]
where \( w_s \) is the weight of the solid.
Step 3: Weight of solid (\( w_s \))
\[
w_s = V_s \cdot G_s \cdot \gamma_w = \frac{V_t}{1+e} \cdot G_s \cdot \gamma_w
\]
where:
- \( V_t = 5 \, \text{m}^3 \) is the total volume.
- \( \gamma_w = 9.81 \, \text{kN/m}^3 \) is the unit weight of water.
Step 4: Substituting the values
\[
0.2247 - 0.15 = \frac{w}{\frac{5}{1.6} \times 2.67 \times 9.81}
\]
Solving for \( w \):
\[
w = 0.0747 \times 3.125 \times 2.67 \times 9.81 = 6 \, \text{kN}
\]
Thus, the required weight of water is:
\[
\boxed{6} \, \text{kN}
\]