1. Magnetic field inside a uniformly current-carrying wire

For a long straight wire of radius $a$ carrying total current $I$ uniformly distributed over its cross-section:

• For a point inside the wire at distance $r<a$, $B_{\text{inside}}=\frac{\mu_0 I_{\text{enc}}}{2\pi r}$

Since current density is uniform, $I_{\text{enc}}=I\frac{\pi r^2}{\pi a^2}=I\frac{r^2}{a^2}$

So, $B_{\text{inside}}=\frac{\mu_0}{2\pi r}\cdot I\frac{r^2}{a^2}=\frac{\mu_0 I r}{2\pi a^2}$

2. Magnetic field at $r=\dfrac{a}{3}$

Since $\dfrac{a}{3}<a$, this point lies inside the wire.

Thus, $B_1=\frac{\mu_0 I}{2\pi a^2}\cdot \frac{a}{3}=\frac{\mu_0 I}{6\pi a}$

3. Magnetic field outside the wire

For a point outside the wire at distance $r>a$, $B_{\text{outside}}=\frac{\mu_0 I}{2\pi r}$

4. Magnetic field at $r=2a$

Since $2a>a$, this point lies outside the wire.

So, $B_2=\frac{\mu_0 I}{2\pi (2a)}=\frac{\mu_0 I}{4\pi a}$

5. Required ratio

$\frac{B_1}{B_2}=\frac{\frac{\mu_0 I}{6\pi a}}{\frac{\mu_0 I}{4\pi a}}=\frac{4}{6}=\frac{2}{3}$

6. Match with options

The ratio is $\boxed{\frac{2}{3}}$ which corresponds to Option D.