The coercivity of a small magnet, where the ferromagnet gets demagnetised is $$3 \times 10^3$$ A/m. The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetised when inside the solenoid is:

We are told that the magnet will be demagnetised when it experiences an opposing magnetic field intensity (also called magnetic field strength) equal to its coercivity. Mathematically, the required magnetic field intensity is

$$H_c = 3 \times 10^3\ \text{A m}^{-1}.$$

The magnet is placed inside a long solenoid. For an ideal long solenoid, the magnetic field intensity $$H$$ produced at any interior point is related to the current $$I$$ flowing through it by the well-known relation

$$H = n I,$$

where $$n$$ is the number of turns per unit length of the solenoid.

Now we calculate $$n$$. The data given are:

Number of turns $$N = 100$$,   length of the solenoid $$L = 10\ \text{cm}.$$ We first convert the length to metres so that all quantities are in SI units:

$$L = 10\ \text{cm} = 10 \times 10^{-2}\ \text{m} = 0.10\ \text{m}.$$

Hence, the turn density is

$$n = \frac{N}{L} = \frac{100}{0.10} = 1000\ \text{turns m}^{-1}.$$

To demagnetise the magnet, the solenoid must supply an equal magnetic field intensity, so we set

$$H = H_c.$$

Substituting $$H = nI$$ and the numerical values, we get

$$n I = H_c$$ $$\Longrightarrow I = \frac{H_c}{n} = \frac{3 \times 10^3\ \text{A m}^{-1}}{1000\ \text{m}^{-1}} = 3\ \text{A}.$$

So, the current that has to be passed through the solenoid is

$$I = 3\ \text{A}.$$

Hence, the correct answer is Option C.