Let’s break down the solution step by step:

First, note the given values:

Length of the wire, $$L = 3 \, \text{m}$$

Radius of the wire, $$r = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m}$$

Extension of the wire, $$\Delta L = 0.1 \, \text{mm} = 1 \times 10^{-4} \, \text{m}$$

Mass attached, $$m = 50 \, \text{kg}$$

Acceleration due to gravity, $$g = 3\pi \, \text{m/s}^2$$

The force (weight) acting on the wire is:

$$F = mg = 50 \times 3\pi = 150\pi \, \text{N}$$

Next, calculate the cross-sectional area, $$A,$$ of the wire:

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

Young’s modulus, $$E,$$ is given by:

$$E = \frac{FL}{A \Delta L}$$

Substitute the values into the formula:

$$ E = \frac{(150\pi) \times 3}{(9\pi \times 10^{-6}) \times (1 \times 10^{-4})} $$

Simplify step by step:

Multiply in the numerator:

$$150\pi \times 3 = 450\pi$$

Multiply in the denominator:

$$9\pi \times 10^{-6} \times 1 \times 10^{-4} = 9\pi \times 10^{-10}$$

Cancel the common factor $$\pi$$ in the numerator and denominator:

$$ E = \frac{450}{9 \times 10^{-10}} $$

Divide 450 by 9:

$$ E = \frac{450}{9} \times \frac{1}{10^{-10}} = 50 \times 10^{10} = 5 \times 10^{11} \, \text{Nm}^{-2} $$

The problem states that Young's modulus can be expressed as $$P \times 10^{11} \, \text{Nm}^{-2}.$$ Here we have:

$$ P = 5 $$

Thus, the correct option is:

Option D (5).