When a block of wood with a weight of $\text{30 N}$ is completely submerged under water the buoyant force on the block of wood from the water is $\text{50 N}$. When the block is released it floats at the surface. What fraction of the block will then be visible above the surface of the water when the block is floating?

(A) $1/15$
(B) $1/5$
(C) $1/3$
(D) $2/5$
(E) $3/5$
1. **Understanding the Problem:**
   - The weight of the block of wood is \( W = 30 \, \text{N} \).
   - The buoyant force when the block is completely submerged is \( F_b = 50 \, \text{N} \).
   - When the block is floating, the buoyant force equals the weight of the block.

2. **Applying Archimedes' Principle:**
   - Archimedes' Principle states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
   - When the block is floating, the buoyant force \( F_b \) must equal the weight of the block \( W \).

3. **Calculating the Fraction Submerged:**
   - Let \( V \) be the total volume of the block.
   - Let \( V_s \) be the volume of the block submerged in water when it is floating.
   - The weight of the water displaced by the submerged part of the block is equal to the buoyant force, which is equal to the weight of the block:
     \[
     \rho_{\text{water}} g V_s = W
     \]
     where \( \rho_{\text{water}} \) is the density of water and \( g \) is the acceleration due to gravity.
   - Given \( W = 30 \, \text{N} \), we have:
     \[
     \rho_{\text{water}} g V_s = 30 \, \text{N}
     \]

4. **Relating Submerged Volume to Total Volume:**
   - When the block is completely submerged, the buoyant force is \( 50 \, \text{N} \):
     \[
     \rho_{\text{water}} g V = 50 \, \text{N}
     \]
   - Therefore, the total volume \( V \) of the block is:
     \[
     V = \frac{50 \, \text{N}}{\rho_{\text{water}} g}
     \]
   - The volume submerged \( V_s \) when the block is floating is:
     \[
     V_s = \frac{30 \, \text{N}}{\rho_{\text{water}} g}
     \]

5. **Finding the Fraction Submerged:**
   - The fraction of the block submerged is:
     \[
     \frac{V_s}{V} = \frac{\frac{30 \, \text{N}}{\rho_{\text{water}} g}}{\frac{50 \, \text{N}}{\rho_{\text{water}} g}} = \frac{30}{50} = \frac{3}{5}
     \]

6. **Finding the Fraction Visible:**
   - The fraction of the block visible above the surface is:
     \[
     1 - \frac{V_s}{V} = 1 - \frac{3}{5} = \frac{2}{5}
     \]

Conclusion:
\[
\boxed{\frac{2}{5}}
\]