Step-by-step Solution:

1. Identify the given information:

   • The edges of the parallelopiped are of unit length. Let these edges be represented by the vectors $\overrightarrow a$, $\overrightarrow b$, and $\overrightarrow c$.
   • Since they are unit vectors, we have $|\overrightarrow a| = |\overrightarrow b| = |\overrightarrow c| = 1$.
   • The dot products between these non-coplanar unit vectors are given: $\overrightarrow a \cdot \overrightarrow b = \overrightarrow b \cdot \overrightarrow c = \overrightarrow c \cdot \overrightarrow a = {1 \over 2}$.

2. Recall the formula for the volume of a parallelopiped: The volume V of a parallelopiped with adjacent edges $\overrightarrow a$, $\overrightarrow b$, and $\overrightarrow c$ is given by the magnitude of the scalar triple product: $V = |[\overrightarrow a \,\overrightarrow b \,\overrightarrow c]| = |\overrightarrow a \cdot (\overrightarrow b \times \overrightarrow c)|$

3. Use the determinant property of the scalar triple product: It is often easier to work with the square of the volume, $V^2$. The square of the scalar triple product can be expressed as a determinant involving the dot products of the vectors: $V^2 = [\overrightarrow a \,\overrightarrow b \,\overrightarrow c]^2 = \begin{vmatrix} \overrightarrow a \cdot \overrightarrow a & \overrightarrow a \cdot \overrightarrow b & \overrightarrow a \cdot \overrightarrow c \\ \overrightarrow b \cdot \overrightarrow a & \overrightarrow b \cdot \overrightarrow b & \overrightarrow b \cdot \overrightarrow c \\ \overrightarrow c \cdot \overrightarrow a & \overrightarrow c \cdot \overrightarrow b & \overrightarrow c \cdot \overrightarrow c \end{vmatrix}$

4. Substitute the given values into the determinant:

   • From $|\overrightarrow a| = |\overrightarrow b| = |\overrightarrow c| = 1$, we get:
     • $\overrightarrow a \cdot \overrightarrow a = |\overrightarrow a|^2 = 1^2 = 1$
     • $\overrightarrow b \cdot \overrightarrow b = |\overrightarrow b|^2 = 1^2 = 1$
     • $\overrightarrow c \cdot \overrightarrow c = |\overrightarrow c|^2 = 1^2 = 1$
   • The cross dot products are given: $\overrightarrow a \cdot \overrightarrow b = \overrightarrow b \cdot \overrightarrow c = \overrightarrow c \cdot \overrightarrow a = {1 \over 2}$.
   • Since the dot product is commutative, $\overrightarrow b \cdot \overrightarrow a = \overrightarrow a \cdot \overrightarrow b$, etc.

   Substituting these values into the determinant for $V^2$: $V^2 = \begin{vmatrix} 1 & 1/2 & 1/2 \\ 1/2 & 1 & 1/2 \\ 1/2 & 1/2 & 1 \end{vmatrix}$

5. Calculate the determinant: We expand the determinant along the first row: $V^2 = 1 \begin{vmatrix} 1 & 1/2 \\ 1/2 & 1 \end{vmatrix} - {1 \over 2} \begin{vmatrix} 1/2 & 1/2 \\ 1/2 & 1 \end{vmatrix} + {1 \over 2} \begin{vmatrix} 1/2 & 1 \\ 1/2 & 1/2 \end{vmatrix}$ $V^2 = 1(1 \cdot 1 - {1 \over 2} \cdot {1 \over 2}) - {1 \over 2}({1 \over 2} \cdot 1 - {1 \over 2} \cdot {1 \over 2}) + {1 \over 2}({1 \over 2} \cdot {1 \over 2} - 1 \cdot {1 \over 2})$ $V^2 = 1(1 - {1 \over 4}) - {1 \over 2}({1 \over 2} - {1 \over 4}) + {1 \over 2}({1 \over 4} - {1 \over 2})$ $V^2 = 1({3 \over 4}) - {1 \over 2}({1 \over 4}) + {1 \over 2}(-{1 \over 4})$ $V^2 = {3 \over 4} - {1 \over 8} - {1 \over 8}$ $V^2 = {3 \over 4} - {2 \over 8} = {3 \over 4} - {1 \over 4}$ $V^2 = {2 \over 4} = {1 \over 2}$

6. Find the volume V: Since $V^2 = {1 \over 2}$, the volume V is the positive square root: $V = \sqrt{{1 \over 2}} = {1 \over \sqrt{2}}$

7. Conclusion: The volume of the parallelopiped is $1/\sqrt{2}$. This corresponds to option A.