A sphere of radius R have volume charge density given as:
`rho( r) = Kr` for `r le R`
= 0 for `r gt R`
If electric field at distance `R/2` from centre is
`(KR^(2))/(2Nepsilon_(0))`. Then Nwill be.

Explanation

To solve the problem, we will use Gauss's law to find the electric field at a distance \( R/2 \) from the center of the sphere with a volume charge density given by \( \rho(r) = Kr \) for \( r \leq R \).

### Step-by-step Solution:

1. **Identify the Charge Density**:
The charge density inside the sphere is given by:
\[ \rho(r) = Kr \quad \text{for } r \leq R \]
and \( \rho(r) = 0 \) for \( r > R \).

2. **Choose a Gaussian Surface**:
We will consider a spherical Gaussian surface of radius \( r = \frac{R}{2} \) (inside the sphere).

3. **Calculate the Enclosed Charge**:
To find the electric field using Gauss's law, we first need to calculate the total charge enclosed by the Gaussian surface. The charge \( dq \) in a thin spherical shell of radius \( r \) and thickness \( dr \) is given by:
\[ dq = \rho(r) \cdot dV = \rho(r) \cdot 4\pi r^2 dr \]
Substituting \( \rho(r) = Kr \):
\[ dq = Kr \cdot 4\pi r^2 dr = 4\pi K r^3 dr \]

4. **Integrate to Find Total Charge Enclosed**:
The total charge \( Q_{\text{enc}} \) enclosed by the Gaussian surface from \( 0 \) to \( \frac{R}{2} \) is:
\[ Q_{\text{enc}} = \int_0^{R/2} 4\pi K r^3 dr \]
Evaluating the integral:
\[ Q_{\text{enc}} = 4\pi K \left[ \frac{r^4}{4} \right]_0^{R/2} = 4\pi K \cdot \frac{(R/2)^4}{4} = \frac{\pi K R^4}{4} \]

5. **Apply Gauss's Law**:
According to Gauss's law:
\[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \]
The left side simplifies to:
\[ E \cdot 4\pi \left(\frac{R}{2}\right)^2 = E \cdot \pi R^2 \]
Therefore, we have:
\[ E \cdot \pi R^2 = \frac{\frac{\pi K R^4}{4}}{\epsilon_0} \]

6. **Solve for Electric Field \( E \)**:
Rearranging gives:
\[ E = \frac{K R^4}{4 \epsilon_0 R^2} = \frac{K R^2}{4 \epsilon_0} \]

7. **Compare with Given Electric Field**:
We are given that the electric field at distance \( R/2 \) is:
\[ E = \frac{K R^2}{2 N \epsilon_0} \]
Setting the two expressions for \( E \) equal:
\[ \frac{K R^2}{4 \epsilon_0} = \frac{K R^2}{2 N \epsilon_0} \]

8. **Solve for \( N \)**:
Cancel \( K R^2 \) and \( \epsilon_0 \) from both sides:
\[ \frac{1}{4} = \frac{1}{2N} \]
Cross-multiplying gives:
\[ 2N = 4 \implies N = 2 \]

### Final Answer:
\[ N = 2 \]