For spherical charge distribution gives as .
`{{:(,rho = rho_(0)(1-(r)/(3)),"when" r le 3m),(,rho = 0 ,"when" r gt 3m):}`
( where x is the distance form the centre of spherical charge distribution)
The electric field intensity is maximum for the value of r=_______m.

Explanation

To find the value of \( r \) for which the electric field intensity is maximum for the given spherical charge distribution, we can follow these steps:

### Step 1: Understand the Charge Distribution
The charge density \( \rho \) is given by:
\[ \rho = \rho_0 \left(1 - \frac{r}{3}\right) \quad \text{for } r \leq 3 \, \text{m} \]
\[ \rho = 0 \quad \text{for } r > 3 \, \text{m} \]

### Step 2: Use Gauss's Law
According to Gauss's Law, the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
For a spherical surface of radius \( r \), the electric field \( E \) is uniform over the surface, so:
\[ \Phi_E = E \cdot 4\pi r^2 \]

### Step 3: Calculate the Enclosed Charge \( Q_{\text{enc}} \)
To find \( Q_{\text{enc}} \), we need to integrate the charge density over the volume of the sphere:
\[ Q_{\text{enc}} = \int_0^r \rho \, dV = \int_0^r \rho_0 \left(1 - \frac{r'}{3}\right) \cdot 4\pi (r')^2 \, dr' \]
Where \( dV = 4\pi (r')^2 dr' \).

### Step 4: Perform the Integration
Substituting \( \rho \) into the integral:
\[ Q_{\text{enc}} = 4\pi \rho_0 \int_0^r \left(1 - \frac{r'}{3}\right) (r')^2 \, dr' \]
This can be split into two integrals:
\[ Q_{\text{enc}} = 4\pi \rho_0 \left( \int_0^r (r')^2 \, dr' - \frac{1}{3} \int_0^r (r')^3 \, dr' \right) \]
Calculating these integrals:
\[ \int_0^r (r')^2 \, dr' = \frac{r^3}{3}, \quad \int_0^r (r')^3 \, dr' = \frac{r^4}{4} \]
Thus,
\[ Q_{\text{enc}} = 4\pi \rho_0 \left( \frac{r^3}{3} - \frac{1}{3} \cdot \frac{r^4}{4} \right) = 4\pi \rho_0 \left( \frac{r^3}{3} - \frac{r^4}{12} \right) \]

### Step 5: Substitute \( Q_{\text{enc}} \) into Gauss's Law
Now substitute \( Q_{\text{enc}} \) back into Gauss's Law:
\[ E \cdot 4\pi r^2 = \frac{4\pi \rho_0 \left( \frac{r^3}{3} - \frac{r^4}{12} \right)}{\epsilon_0} \]
Cancelling \( 4\pi \) from both sides:
\[ E r^2 = \frac{\rho_0}{\epsilon_0} \left( \frac{r^3}{3} - \frac{r^4}{12} \right) \]

### Step 6: Solve for Electric Field \( E \)
Rearranging gives:
\[ E = \frac{\rho_0}{\epsilon_0} \left( \frac{r}{3} - \frac{r^2}{12} \right) \]

### Step 7: Find Maximum Electric Field
To find the maximum electric field, differentiate \( E \) with respect to \( r \) and set the derivative to zero:
\[ \frac{dE}{dr} = \frac{\rho_0}{\epsilon_0} \left( \frac{1}{3} - \frac{r}{6} \right) = 0 \]
Solving for \( r \):
\[ \frac{1}{3} = \frac{r}{6} \implies r = 2 \, \text{m} \]

### Final Answer
The electric field intensity is maximum at:
\[ \boxed{2 \, \text{m}} \]