Using KCL, KVL and Ohm’s Law find I0 in the circuit in Fig.1.1
First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course).

Well, we are trying to analyze the following circuit:

When we use and apply KCL, we can write the following set of equations:
$$
\begin{cases}
\text{I}_\text{a}=\text{I}_1+\text{I}_5\\
\\
\text{I}_5=\text{n}\cdot\text{I}_2+\text{I}_4\\
\\
\text{I}_1=\text{I}_2+\text{I}_3\\
\\
0=\text{n}\cdot\text{I}_2+\text{I}_4+\text{I}_6\\
\\
\text{I}_6=\text{I}_3+\text{I}_7\\
\\
\text{I}_2=\text{I}_\text{a}+\text{I}_7
\end{cases}\tag1
$$
When we use and apply Ohm's law, we can write the following set of equations:
$$
\begin{cases}
\text{I}_1=\frac{\text{V}_1-\text{V}_2}{\text{R}_1}\\
\\
\text{I}_2=\frac{\text{V}_2}{\text{R}_2}\\
\\
\text{I}_3=\frac{\text{V}_2}{\text{R}_3}\\
\\
\text{I}_4=\frac{\text{V}_1}{\text{R}_4}
\end{cases}\tag2
$$
Substitute \$(2)\$ into \$(1)\$, in order to get:
$$
\begin{cases}
\text{I}_\text{a}=\frac{\text{V}_1-\text{V}_2}{\text{R}_1}+\text{I}_5\\
\\
\text{I}_5=\text{n}\cdot\frac{\text{V}_2}{\text{R}_2}+\frac{\text{V}_1}{\text{R}_4}\\
\\
\frac{\text{V}_1-\text{V}_2}{\text{R}_1}=\frac{\text{V}_2}{\text{R}_2}+\frac{\text{V}_2}{\text{R}_3}\\
\\
0=\text{n}\cdot\frac{\text{V}_2}{\text{R}_2}+\frac{\text{V}_1}{\text{R}_4}+\text{I}_6\\
\\
\text{I}_6=\frac{\text{V}_2}{\text{R}_3}+\text{I}_7\\
\\
\frac{\text{V}_2}{\text{R}_2}=\text{I}_\text{a}+\text{I}_7
\end{cases}\tag3
$$
Solving for \$\text{I}_1\$, gives:
$$\text{I}_4=\frac{\text{I}_\text{a}\left(\text{R}_2\text{R}_3+\text{R}_1\left(\text{R}_2+\text{R}_3\right)\right)}{\text{R}_1\left(\text{R}_2+\text{R}_3\right)+\left(1+\text{n}\right)\text{R}_3\text{R}_4+\text{R}_2\left(\text{R}_3+\text{R}_4\right)}\tag4$$
Where I used the following Mathematica-code:
In[1]:=Clear[&quot;Global`*&quot;];
FullSimplify[
 Solve[{Ia == I1 + I5, I5 == n*I2 + I4, I1 == I2 + I3, 
   0 == n*I2 + I4 + I6, I6 == I3 + I7, I2 == Ia + I7, 
   I1 == (V1 - V2)/R1, I2 == V2/R2, I3 == V2/R3, I4 == V1/R4}, {I1, 
   I2, I3, I4, I5, I6, I7, V1, V2}]]

Out[1]={{I1 -&gt; (Ia (R2 + R3) R4)/(
   R1 (R2 + R3) + (1 + n) R3 R4 + R2 (R3 + R4)), 
  I2 -&gt; (Ia R3 R4)/(R1 (R2 + R3) + (1 + n) R3 R4 + R2 (R3 + R4)), 
  I3 -&gt; (Ia R2 R4)/(R1 (R2 + R3) + (1 + n) R3 R4 + R2 (R3 + R4)), 
  I4 -&gt; (Ia (R2 R3 + R1 (R2 + R3)))/(
   R1 (R2 + R3) + (1 + n) R3 R4 + R2 (R3 + R4)), 
  I5 -&gt; Ia - (Ia (R2 + R3) R4)/(
    R1 (R2 + R3) + (1 + n) R3 R4 + R2 (R3 + R4)), 
  I6 -&gt; -((Ia (R1 (R2 + R3) + R3 (R2 + n R4)))/(
    R1 (R2 + R3) + (1 + n) R3 R4 + R2 (R3 + R4))), 
  I7 -&gt; Ia (-1 + (R3 R4)/(
      R2 R3 + R1 (R2 + R3) + (R2 + R3 + n R3) R4)), 
  V1 -&gt; (Ia (R2 R3 + R1 (R2 + R3)) R4)/(
   R1 (R2 + R3) + (1 + n) R3 R4 + R2 (R3 + R4)), 
  V2 -&gt; (Ia R2 R3 R4)/(R1 (R2 + R3) + (1 + n) R3 R4 + R2 (R3 + R4))}}

Using your values, we get:
$$\text{I}_4=\frac{9}{3250}\approx0.00276923\space\text{A}\tag5$$
The code gives the rest of the answers:
In[2]:=Clear[&quot;Global`*&quot;];
Ia = 6*10^(-3);
n = 2;
R1 = 4*1000;
R2 = 3*1000;
R3 = 6*1000;
R4 = 3*1000;
FullSimplify[
 Solve[{Ia == I1 + I5, I5 == n*I2 + I4, I1 == I2 + I3, 
   0 == n*I2 + I4 + I6, I6 == I3 + I7, I2 == Ia + I7, 
   I1 == (V1 - V2)/R1, I2 == V2/R2, I3 == V2/R3, I4 == V1/R4}, {I1, 
   I2, I3, I4, I5, I6, I7, V1, V2}]]

Out[2]={{I1 -&gt; 9/6500, I2 -&gt; 3/3250, I3 -&gt; 3/6500, I4 -&gt; 9/3250, I5 -&gt; 3/650,
   I6 -&gt; -(3/650), I7 -&gt; -(33/6500), V1 -&gt; 108/13, V2 -&gt; 36/13}}

In[3]:=N[%2]

Out[3]={{I1 -&gt; 0.00138462, I2 -&gt; 0.000923077, I3 -&gt; 0.000461538, 
  I4 -&gt; 0.00276923, I5 -&gt; 0.00461538, I6 -&gt; -0.00461538, 
  I7 -&gt; -0.00507692, V1 -&gt; 8.30769, V2 -&gt; 2.76923}}