Step 1 :

Calculation of $V_{Th}$

From the given figure , in order to find  $V_{Th}$ let us remove the  resistor of $10 \Omega$ from the circuit.

$\therefore V_{x}=-25(5I_1)$

$\therefore V_x=-125I_1$........(1)

Applying KVL to the mesh 1,

$12-1000I_1-2(-125I_1)=0$

$\therefore I_1=0.016A$

Putting the value of $I_1$in equation (1),

$\therefore V_x=-125I_1=-125(0.016)=-2V$

Writing the $V_{Th}$ equation ,

$V_{Th}=V_x=-2V$

Step 2:

Calculation of  $I_N$ 

In order to find the norton current, let us short circuit the $10\Omega$ resistor and current flowing through this resistor is $I_N$.

By short circuiting , $V_x=0$

The dependent source of $2V$ depends on the controlling variable $V_x$ .When  $V_x=0$ , the dependent sources vanishes,i.e. $2V_x=0$.

$\therefore I_1=\dfrac{12}{1000}=0.012A$

$I_N=-5I_1=-5(0.012)=-0.06A$

Step 3 :

Calculation of $R_N$

$\therefore R_N=\dfrac{V_{Th}}{I_N}=\dfrac{-2}{-0.06}=33.33 \Omega$

Step 4:

Calculation of  $I_L$ 

$I_L$ is the current flowing through $10\Omega$ resistor of nortons equivalent circuit with current source of -0.06A and $R_N=33.33 \Omega$ in the circuit.

$I_L=-0.06\times \dfrac{33.33}{33.33+10}$

$\therefore I_L=-0.046A$

Hence, the current flowing through $10\Omega $ resistor by Norton's theorem is -0.046A.