B

In Young's experiment, the value of Young's modulus (Y) is given by the formula:
Y = \frac{F/A}{ΔL/L}
Where:
- F is the force applied
- A is the cross-sectional area
- ΔL is the change in length
- L is the original length

If the length of the wire (L) is doubled, the new length becomes 2L. If the radius (r) is also doubled, the new radius becomes 2r.
The cross-sectional area (A) of the wire is given by the formula:
A = \pi r^2
Therefore, if the radius is doubled, the new area becomes:
A' = \pi (2r)^2 = \pi (4r^2) = 4A

The new Young's modulus after these changes can be expressed as:
Y' = \frac{F/A'}{ΔL/L'}
So, replacing A' and L' in the formula, we get:
Y' = \frac{F/(4A)}{ΔL/(2L)} = \frac{F}{4A} \cdot \frac{2L}{ΔL} = \frac{1}{2} \cdot \frac{F/A}{ΔL/L} = \frac{1}{2}Y

Therefore, Y becomes four times the original, i.e., 4Y when taking into account the change in area and length. The value of Young’s modulus thus becomes 4 times.
Therefore, the correct answer is option B: 4 times.