The fractional change in the magnetic field intensity at a distance $$r$$ from centre on the axis of current carrying coil of radius $$a$$ to the magnetic field intensity at the centre of the same coil is: (Take $$r < a$$)

We need the magnetic field produced by a circular coil carrying a steady current. The standard expression for the magnitude of the magnetic field on the axis of a single loop of radius $$a$$, at a point whose distance from the centre measured along the axis is $$r$$, is

$$B(r)=\frac{\mu_{0}\,I\,a^{2}}{2\left(a^{2}+r^{2}\right)^{3/2}}.$$

(This formula can be derived from the Biot-Savart law, but we take it here as a known result.) At the centre of the same coil we simply place $$r=0$$, giving

$$B_{\text{centre}}=B(0)=\frac{\mu_{0}\,I\,a^{2}}{2\left(a^{2}+0\right)^{3/2}} =\frac{\mu_{0}\,I\,a^{2}}{2a^{3}} =\frac{\mu_{0}\,I}{2a}.$$

We now form the ratio of the field at the off-axis point to the field at the centre:

$$\frac{B(r)}{B_{\text{centre}}} =\frac{\dfrac{\mu_{0}I a^{2}}{2\left(a^{2}+r^{2}\right)^{3/2}}} {\dfrac{\mu_{0}I}{2a}} =\frac{a^{3}}{\left(a^{2}+r^{2}\right)^{3/2}} =\frac{1}{\left(1+\dfrac{r^{2}}{a^{2}}\right)^{3/2}}.$$

To find the fractional change we write

$$\text{Fractional change}=\frac{B(r)-B_{\text{centre}}}{B_{\text{centre}}} =\frac{B(r)}{B_{\text{centre}}}-1 =\left(1+\frac{r^{2}}{a^{2}}\right)^{-3/2}-1.$$

The question specifies $$r<a$$, so the quantity $$\dfrac{r^{2}}{a^{2}}$$ is small. We therefore expand the power using the binomial approximation. For a small $$x$$, the expansion

$$(1+x)^{n}\;\approx\;1+nx$$

is adequate to first order. Here $$n=-\dfrac{3}{2}$$ and $$x=\dfrac{r^{2}}{a^{2}}$$, so

$$\left(1+\frac{r^{2}}{a^{2}}\right)^{-3/2} \approx 1-\frac{3}{2}\frac{r^{2}}{a^{2}}.$$

Substituting this approximation into the expression for the fractional change gives

$$\text{Fractional change} \approx\Bigl[1-\frac{3}{2}\frac{r^{2}}{a^{2}}\Bigr]-1 =-\frac{3}{2}\frac{r^{2}}{a^{2}}.$$

The negative sign merely tells us that the magnetic field at the point on the axis is smaller than the field at the centre. The magnitude of the fractional change is therefore

$$\left|\frac{B(r)-B_{\text{centre}}}{B_{\text{centre}}}\right| =\frac{3}{2}\frac{r^{2}}{a^{2}}.$$

Among the given options, this matches

$$\frac{3}{2}\frac{r^{2}}{a^{2}}.$$

Hence, the correct answer is Option C.