A galvanometer coil has 500 turns and each turn has an average area of $$3 \times 10^{-4}$$ m$$^2$$. If a torque of 1.5 Nm is required to keep this coil parallel to a magnetic field when a current of 0.5A is flowing through it, the strength of the field (in T) is __________.

We recall the torque formula for a current-carrying coil placed in a uniform magnetic field. The magnitude of the torque is given by

$$\tau = N I A B \sin\theta,$$

where $$N$$ is the number of turns, $$I$$ is the current, $$A$$ is the area of each turn, $$B$$ is the magnetic-field strength, and $$\theta$$ is the angle between the magnetic-field direction and the normal to the plane of the coil.

Now the coil is stated to be parallel to the magnetic field. This makes the normal to the coil perpendicular to the field, giving $$\theta = 90^\circ$$ and hence $$\sin\theta = 1$$. Therefore the expression for torque simplifies to

$$\tau = N I A B.$$

We substitute the given numerical values. We have

$$N = 500, \qquad I = 0.5\ \text{A}, \qquad A = 3 \times 10^{-4}\ \text{m}^2, \qquad \tau = 1.5\ \text{N\,m}.$$

Putting these into the simplified torque formula gives

$$1.5 = 500 \times 0.5 \times \left(3 \times 10^{-4}\right) \times B.$$

First evaluate the product in the denominator step by step:

$$500 \times 0.5 = 250,$$

$$250 \times 3 = 750,$$

$$750 \times 10^{-4} = 7.5 \times 10^{-2} = 0.075.$$

So we have

$$1.5 = 0.075 \, B.$$

Solving for $$B$$, we divide both sides by 0.075:

$$B = \frac{1.5}{0.075} = 20.$$

Thus the required magnetic-field strength is

$$B = 20\ \text{T}.$$

So, the answer is $$20\ \text{T}$$.