1. Given data

• Original length of rubber cord: $L = 42\text{ cm} = 0.42\text{ m}$
• Extension: $x = 20\text{ cm} = 0.20\text{ m}$
• Diameter of cord: $d = 6\text{ mm} = 6 \times 10^{-3}\text{ m}$
• Radius: $r = 3 \times 10^{-3}\text{ m}$
• Mass of stone: $m = 0.02\text{ kg}$
• Speed of stone on release: $v = 20\text{ m s}^{-1}$

We assume the elastic potential energy stored in the stretched rubber becomes the kinetic energy of the stone.

2. Elastic energy stored in the cord

Since the boy stretches the cord by applying a constant force, the force-extension graph is a straight line from $0$ to the final force $F$.

Hence, elastic energy stored is

$U = \frac{1}{2}Fx$

This becomes kinetic energy of the stone:

$\frac{1}{2}Fx = \frac{1}{2}mv^2$

So,

$Fx = mv^2$

$F = \frac{mv^2}{x}$

Substitute values:

$F = \frac{0.02 \times (20)^2}{0.20} = \frac{0.02 \times 400}{0.20} = \frac{8}{20} \times 10 = 40\text{ N}$

So the final stretching force is

$F = 40\text{ N}$

3. Use Young's modulus formula

Young's modulus is

$Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{x/L} = \frac{FL}{Ax}$

Cross-sectional area:

$A = \pi r^2 = \pi (3 \times 10^{-3})^2 = 9\pi \times 10^{-6}\text{ m}^2$

Now,

$Y = \frac{40 \times 0.42}{(9\pi \times 10^{-6}) \times 0.20}$

$Y = \frac{16.8}{1.8\pi \times 10^{-6}}$

Using $\pi \approx 3.14$,

$Y \approx \frac{16.8}{5.652 \times 10^{-6}}$

$Y \approx 2.97 \times 10^6\text{ N m}^{-2}$

Thus,

$Y \approx 3 \times 10^6\text{ N m}^{-2}$

4. Choose the closest option

The value is of the order of

$10^6\text{ N m}^{-2}$

So the closest option is:

Option B: $10^6\text{ N m}^{-2}$

5. Verification with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.