Explanation:

A helical spring is a mechanical device used to store energy and provide resistance to a force. It is made up of a wire wound in the form of a helix, with a spring constant (k) that determines its stiffness.

Given:
- The spring constant of the original spring is k.
- The wire diameter, spring diameter, and the number of coils are all doubled.

Effect of doubling the wire diameter:
- Doubling the wire diameter will increase the cross-sectional area of the wire, resulting in a decrease in the effective spring constant.
- The spring constant is inversely proportional to the cross-sectional area of the wire.
- Therefore, the new spring constant after doubling the wire diameter will be k/2.

Effect of doubling the spring diameter:
- Doubling the spring diameter will increase the length of the spring, resulting in an increase in the spring constant.
- The spring constant is directly proportional to the length of the spring.
- Therefore, the new spring constant after doubling the spring diameter will be 2k.

Effect of doubling the number of coils:
- Doubling the number of coils will increase the length of the spring, resulting in an increase in the spring constant.
- The spring constant is directly proportional to the length of the spring.
- Therefore, the new spring constant after doubling the number of coils will be 2k.

Overall effect:
- When all three parameters (wire diameter, spring diameter, and number of coils) are doubled, the effects on the spring constant are as follows:
- Wire diameter: k/2
- Spring diameter: 2k
- Number of coils: 2k
- The overall effect on the spring constant is the multiplication of these individual effects.
- Therefore, the new spring constant after doubling all three parameters will be (k/2) * (2k) * (2k) = 4k^2.

Conclusion:
- The spring constant of the new spring after doubling the wire diameter, spring diameter, and number of coils is 4k^2.
- The correct answer is option 'C', which states that the spring constant becomes k/2.