1. Given data

• Coercivity of magnet: $H_c = 5 \times 10^3\ \text{A/m}$
• Length of solenoid: $l = 30\ \text{cm} = 0.30\ \text{m}$
• Number of turns: $N = 150$

We need the current required so that the magnetic field intensity produced by the solenoid equals the coercivity of the magnet.

2. Magnetic field intensity inside a solenoid

For a solenoid,

$H = \frac{NI}{l}$

To demagnetise the magnet,

$H = H_c$

So,

$\frac{NI}{l} = H_c$

3. Substitute the values

$\frac{150\, I}{0.30} = 5 \times 10^3$

$500 I = 5000$

$I = \frac{5000}{500} = 10$

4. Final answer

The required current is

$\boxed{10\ \text{A}}$