A long wire in the shape of a cylindrical shell of inner and outer radii `R/2` and R respectively carries a steady current I in a direction parallel to its length. This current is uniformly distributed across the area of cross-section of the wire. The magnetic field intensity at a point at a distance `(3R)/4` from the axis of the wire is `m((mu_(0)I)/(pi R))`.. The value of m is _________.

Explanation

To solve the problem, we need to find the magnetic field intensity at a distance of \( \frac{3R}{4} \) from the axis of a cylindrical shell carrying a steady current \( I \). The cylindrical shell has inner and outer radii of \( \frac{R}{2} \) and \( R \), respectively.

### Step-by-Step Solution:

1. **Identify the Geometry**:
- The cylindrical shell has inner radius \( r_1 = \frac{R}{2} \) and outer radius \( r_2 = R \).
- We need to calculate the magnetic field at a distance \( r = \frac{3R}{4} \) from the axis of the wire.

2. **Calculate the Cross-Sectional Area**:
- The area of the cross-section of the cylindrical shell is given by:
\[ A = \pi r_2^2 - \pi r_1^2 = \pi R^2 - \pi \left(\frac{R}{2}\right)^2 = \pi R^2 - \frac{\pi R^2}{4} = \frac{3\pi R^2}{4} \]

3. **Determine the Current Density**:
- The current density \( J \) (current per unit area) is given by:
\[ J = \frac{I}{A} = \frac{I}{\frac{3\pi R^2}{4}} = \frac{4I}{3\pi R^2} \]

4. **Use Ampere's Circuital Law**:
- According to Ampere's Law, the magnetic field \( B \) around a long straight conductor is given by:
\[ B \cdot 2\pi r = \mu_0 I_{\text{enc}} \]
- Here, \( I_{\text{enc}} \) is the current enclosed by the Amperian loop of radius \( r = \frac{3R}{4} \).

5. **Calculate the Enclosed Current**:
- The enclosed current \( I_{\text{enc}} \) for a radius \( r = \frac{3R}{4} \) is calculated by considering the area of the shell from \( r_1 \) to \( r \):
\[ A_{\text{enc}} = \pi r^2 - \pi r_1^2 = \pi \left(\frac{3R}{4}\right)^2 - \pi \left(\frac{R}{2}\right)^2 \]
\[ = \pi \left(\frac{9R^2}{16} - \frac{R^2}{4}\right) = \pi \left(\frac{9R^2}{16} - \frac{4R^2}{16}\right) = \pi \left(\frac{5R^2}{16}\right) \]
- The enclosed current is then:
\[ I_{\text{enc}} = J \cdot A_{\text{enc}} = \frac{4I}{3\pi R^2} \cdot \pi \left(\frac{5R^2}{16}\right) = \frac{4I \cdot 5}{3 \cdot 16} = \frac{20I}{48} = \frac{5I}{12} \]

6. **Substitute into Ampere's Law**:
- Now substituting \( I_{\text{enc}} \) back into Ampere's Law:
\[ B \cdot 2\pi \left(\frac{3R}{4}\right) = \mu_0 \cdot \frac{5I}{12} \]
- Simplifying gives:
\[ B \cdot \frac{3\pi R}{2} = \mu_0 \cdot \frac{5I}{12} \]
- Thus,
\[ B = \frac{\mu_0 \cdot 5I}{12} \cdot \frac{2}{3\pi R} = \frac{5\mu_0 I}{18\pi R} \]

7. **Relate to Given Expression**:
- The problem states that \( B = m \cdot \frac{\mu_0 I}{\pi R} \).
- Therefore, equating gives:
\[ m = \frac{5}{18} \]

### Final Answer:
The value of \( m \) is \( \frac{5}{18} \).