A paramagnetic substance in the form of a cube with sides 1 cm has a magnetic dipole moment of $$20 \times 10^{-6}$$ J/T when a magnetic intensity of $$60 \times 10^3$$ A/m is applied. Its magnetic susceptibility is:

We begin by recalling that the magnetic moment of a uniformly magnetised body is related to its magnetisation by

$$\vec M \;=\; \frac{\text{total magnetic moment}}{\text{volume of the body}}\;.$$

Here the substance is a cube whose edge length is given as $$1\ \text{cm}$$. Converting this length into SI units, we have

$$1\ \text{cm}=1\times10^{-2}\ \text{m}=0.01\ \text{m}\,.$$

Now the volume of a cube is the cube of its side, so

$$V=(0.01\ \text{m})^{3}=0.01^{3}\ \text{m}^{3}=1\times10^{-6}\ \text{m}^{3}\,.$$

The total magnetic dipole moment of the cube is given as $$20\times10^{-6}\ \text{J/T}$$. Remembering that $$1\ \text{J/T}=1\ \text{A·m}^{2}$$, we can write

$$\mu =20\times10^{-6}\ \text{A·m}^{2}\,.$$

Substituting the values of $$\mu$$ and $$V$$ in the definition of magnetisation, we get

$$M=\frac{\mu}{V}=\frac{20\times10^{-6}\ \text{A·m}^{2}}{1\times10^{-6}\ \text{m}^{3}}=20\ \text{A/m}\,.$$

Next, magnetic susceptibility $$\chi_{m}$$ is defined as the ratio of magnetisation $$M$$ to the applied magnetic intensity $$H$$, that is,

$$\chi_{m}=\frac{M}{H}\,.$$

The applied magnetic intensity is given as $$60\times10^{3}\ \text{A/m}=60000\ \text{A/m}$$. Hence, substituting the known values,

$$\chi_{m}=\frac{20\ \text{A/m}}{60000\ \text{A/m}} =\frac{20}{60000} =\frac{2}{6000} =\frac{1}{3000} =3.33\times10^{-4}\,.$$

So we obtain

$$\chi_{m}\;\approx\;3.3\times10^{-4}\,.$$

Hence, the correct answer is Option D.