The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is $$B_1$$. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is $$B_2$$. The ratio $$\frac{B_1}{B_2}$$ is:

For a circular current-carrying loop the magnetic dipole moment is defined by the formula $$m = I\,A$$ where $$I$$ is the current and $$A$$ is the area enclosed by the loop.

We are told that, with the same current $$I$$, the dipole moment is doubled. Starting with an initial dipole moment $$m$$ and a final dipole moment $$2m$$, we write

$$m_1 \;=\; I\,A_1 = m,$$ $$m_2 \;=\; I\,A_2 = 2m.$$

Since the current $$I$$ remains unchanged, we can equate the ratios:

$$\dfrac{A_2}{A_1} = \dfrac{m_2}{m_1} = \dfrac{2m}{m} = 2.$$

Hence the area has become twice its original value, so

$$A_2 = 2A_1.$$

The area of a circle is given by the well-known relation $$A = \pi r^{2}.$$ Writing this for both configurations, we have

$$A_1 = \pi r_1^{2}, \qquad A_2 = \pi r_2^{2}.$$

Substituting $$A_2 = 2A_1$$ gives

$$\pi r_2^{2} = 2 \,(\pi r_1^{2}).$$

Cancelling the common factor $$\pi$$ and solving for $$r_2$$,

$$r_2^{2} = 2 r_1^{2} \;\;\Longrightarrow\;\; r_2 = \sqrt{2}\,r_1.$$

The magnetic field at the centre of a single circular loop is given by the expression

$$B = \dfrac{\mu_0 I}{2r},$$

where $$\mu_0$$ is the permeability of free space and $$r$$ is the radius of the loop. Using this formula for the two cases,

$$B_1 = \dfrac{\mu_0 I}{2 r_1},$$ $$B_2 = \dfrac{\mu_0 I}{2 r_2}.$$

Now we substitute $$r_2 = \sqrt{2}\,r_1$$ into the expression for $$B_2$$:

$$B_2 = \dfrac{\mu_0 I}{2 (\sqrt{2}\,r_1)} = \dfrac{1}{\sqrt{2}}\,\dfrac{\mu_0 I}{2 r_1} = \dfrac{B_1}{\sqrt{2}}.$$

Finally, taking the ratio of the initial to the final field,

$$\dfrac{B_1}{B_2} = \dfrac{B_1}{\,B_1/\sqrt{2}\,} = \sqrt{2}.$$

Hence, the correct answer is Option D.