Let $$B_1$$ be the magnitude of magnetic field at center of a circular coil of radius R carrying current I. Let $$B_2$$ be the magnitude of magnetic field at an axial distance 'x' from the center. For $$x : R = 3 : 4$$, $$\frac{B_2}{B_1}$$ is:

For a circular coil of radius $$R$$ carrying current $$I$$, the standard results for magnetic field are:

• At the centre of the coil (on its plane) $$B_1 = \frac{\mu_0 I}{2R} \quad -(1)$$

• On the axis of the coil at a distance $$x$$ from the centre $$B_2 = \frac{\mu_0 I R^{2}}{2\left(R^{2}+x^{2}\right)^{3/2}} \quad -(2)$$

To obtain the required ratio, divide $$(2)$$ by $$(1)$$:

$$\frac{B_2}{B_1} = \frac{\mu_0 I R^{2}}{2\left(R^{2}+x^{2}\right)^{3/2}} \times \frac{2R}{\mu_0 I} = \frac{R^{3}}{\left(R^{2}+x^{2}\right)^{3/2}} \quad -(3)$$

Rewrite $$(3)$$ more compactly:

$$\frac{B_2}{B_1} = \left(\frac{R^{2}}{R^{2}+x^{2}}\right)^{3/2} \quad -(4)$$

The problem specifies $$x : R = 3 : 4$$, so choose $$x = 3k$$ and $$R = 4k$$ for some constant $$k$$.

Substitute these into $$(4)$$:

$$\frac{B_2}{B_1} = \left( \frac{(4k)^{2}}{(4k)^{2} + (3k)^{2}} \right)^{3/2} = \left( \frac{16k^{2}}{16k^{2} + 9k^{2}} \right)^{3/2} = \left( \frac{16}{25} \right)^{3/2} \quad -(5)$$

Evaluate the power:

$$\left( \frac{16}{25} \right)^{3/2} = \frac{16^{3/2}}{25^{3/2}} = \frac{( \sqrt{16} )^{3}}{( \sqrt{25} )^{3}} = \frac{4^{3}}{5^{3}} = \frac{64}{125}$$

Hence, $$\displaystyle \frac{B_2}{B_1} = \frac{64}{125}$$, or in ratio form $$64 : 125$$.

Therefore, the correct option is Option C (64 : 125).