Solution:

To solve for the magnetic field at a distance of $\frac{5R}{4}$ from the axis of a hollow cylindrical wire carrying current $I$, with inner radius $R$ and outer radius $2R$, we use Ampère's Law. The key steps are:

1. Magnetic Field Inside a Hollow Cylinder

For $R < r < 2R$ (points within the shell of the cylinder):

• Current density $J$ is uniform, given by:

  $$J = \frac{I}{\pi \left((2R)^2 - R^2\right)} = \frac{I}{3\pi R^2}.$$

• The current enclosed within a radius $r$ (where $R < r < 2R$) is:

  $$I_{\text{enc}} = J \cdot \text{area enclosed} = J \cdot \pi (r^2 - R^2).$$Substituting $J$:

  $$I_{\text{enc}} = \frac{I}{3\pi R^2} \cdot \pi (r^2 - R^2) = \frac{I}{3R^2}(r^2 - R^2).$$

• Using Ampère's Law:

  $$B \cdot 2\pi r = \mu_0 I_{\text{enc}},$$ $$B = \frac{\mu_0}{2\pi r} \cdot \frac{I}{3R^2}(r^2 - R^2).$$

2. Magnetic Field at $r = \frac{5R}{4}$

Since $\frac{5R}{4}$ lies within the shell ($R < r < 2R$), substitute $r = \frac{5R}{4}$ into the above equation:

$$B = \frac{\mu_0}{2\pi \cdot \frac{5R}{4}} \cdot \frac{I}{3R^2}\left(\left(\frac{5R}{4}\right)^2 - R^2\right).$$

Simplify:

• $r^2 = \left(\frac{5R}{4}\right)^2 = \frac{25R^2}{16}$,
• $r^2 - R^2 = \frac{25R^2}{16} - R^2 = \frac{25R^2}{16} - \frac{16R^2}{16} = \frac{9R^2}{16}$.

Thus:

$$B = \frac{\mu_0}{2\pi \cdot \frac{5R}{4}} \cdot \frac{I}{3R^2} \cdot \frac{9R^2}{16}.$$

Simplify further:

$$B = \frac{\mu_0}{2\pi} \cdot \frac{4}{5R} \cdot \frac{9I}{48} = \frac{\mu_0 I}{2\pi} \cdot \frac{36}{240R}.$$

Final simplification:

$$B = \frac{3}{40} \frac{\mu_0 I}{\pi R}.$$

Final Answer:

$$\boxed{B = \frac{3}{40} \frac{\mu_0 I}{\pi R}}$$