Step 1: Link adj A to A:
Use \(A \cdot \text{adj } A = |A| I\). Take determinants of both sides for a $3 \times 3$ matrix: $|A| \cdot |\text{adj } A| = |A|^3$.
Step 2: Solve for \(|A|\):
Since $|\text{adj } A| = 25$, we have $25|A| = |A|^3$. If $|A|$ were 0, then $|\text{adj } A|$ would also be 0 for a $3 \times 3$ matrix, which contradicts 25. So $|A| \neq 0$. Divide by $|A|$ to get $|A|^2 = 25$, so $|A| = \pm 5$.
Step 3: Scale by 2:
Multiplying every row of $A$ by 2 multiplies the determinant by 2 three times. So $|2A| = 8|A| = 8(\pm 5)$.
Step 4: Answer:
The value is $\pm 40$, which matches option 4.
Final Answer:
\(|2A| = \pm 40\).
\[ \boxed{\pm 40} \]