Error calculating thevenin's equivalent
There are lots of ways to solve the problem. And since Thevenin's Theorem states little more than:

Any linear circuit containing several voltages and resistances can be
replaced by just one single voltage in series with a single resistance
connected across the load.

You are pretty much free to choose your approach in arriving at that destination.
Norton-Thevenin transformation approach
You chose to use KVL. I prefer KCL. And I'll get to both.
But let's just take this simply, at first, and instead use a set of Norton-Thevenin transformations.



simulate this circuit &ndash; Schematic created using CircuitLab
The above should be obvious (and since you applied it yourself, I know you are aware of it.) On the left the only change I made is to set the \$B\$ node as my reference point, or ground. I then converted the Norton source of \$I_1\$ and \$R_1\$ into a Thevenin source shown on the right. And I simplified the schematic slightly, by dumping the entire voltage source and replacing it with a mere note about its voltage.
This is a really simple transformation from the above:



simulate this circuit
I shouldn't even need to explain that one, above. Easy.
At this point, all you have to do is the following:



simulate this circuit
In going from the left side to the top right schematic above, you should already know how to transform a resistor divider that sits between any voltage pair. But just in case others may not be, I'll write it out: \$V_\text{equiv}=\frac{V_\text{A}\cdot R_\text{B}+V_\text{B}\cdot R_\text{A}}{R_\text{A}+R_\text{B}}\$ and \$R_\text{equiv}=\frac{R_\text{A}\cdot R_\text{B}}{R_\text{A}+R_\text{B}}\$, where \$V_\text{A}\$ and \$R_\text{A}\$ are the source voltage and source resistance from one side and \$V_\text{B}\$ and \$R_\text{B}\$ are the source voltage and source resistance from the opposite side of the divider. (It's worth working out by hand to see why these are correct.)
Well, that was easy.
Nodal analysis
Another approach I like uses KCL instead of KVL. Here, referring back to the first schematic, we are looking first to find \$V_\text{A}\$ (which will be the same as \$V_\text{Y}\$ without a load), and find:
$$\begin{align*}
\frac{V_\text{X}}{R_1}+\frac{V_\text{X}}{R_2}&amp;=3\:\text{A}+\frac{V_\text{Y}}{R_2}\\\\
\frac{V_\text{Y}}{R_2}+\frac{V_\text{Y}}{R_3}&amp;=\frac{V_\text{X}}{R_2}+\frac{12\:\text{V}}{R_3}
\end{align*}$$
The above solves out as \$V_\text{Y}=19.2\:\text{V}\$. Just as we got, earlier.
To find the Thevenin resistance of the source, we simply inject \$1\:\text{A}\$ at node \$A\$ and find out the voltage:
$$\begin{align*}
\frac{V_\text{X}}{R_1}+\frac{V_\text{X}}{R_2}&amp;=3\:\text{A}+\frac{V_\text{Y}}{R_2}\\\\
\frac{V_\text{Y}}{R_2}+\frac{V_\text{Y}}{R_3}&amp;=\frac{V_\text{X}}{R_2}+\frac{12\:\text{V}}{R_3}+1\:\text{A}
\end{align*}$$
Here, we get \$V_\text{Y}=23.2\:\text{V}\$. And from this fact, we know that \$V_\text{A}=V_\text{Y}+1\:\text{A}\cdot R_4=24.2\:\text{V}\$. Therefore, the Thevenin resistance must be \$R_\text{TH}=\frac{24.2\:\text{V}-19.2\:\text{V}}{1\:\text{A}}=5\:\Omega\$.
Once again, same answer using a different approach.
Your approach
Let's follow your KVL approach after your made that first Norton-Thevenin conversion:
$$0\:\text{V}+48\:\text{V}-I\cdot R_1 - I\cdot R_2 - I\cdot R_3= 12\:\text{V}$$
We find that \$I=1\frac{11}{25}\:\text{A}\$ and therefore \$V_\text{Y}=12\:\text{V}+I\cdot R_3=19.2\:\text{V}\$.
Same answer as before for the equivalent source voltage. Nothing new, here.
To get the equivalent source resistance, we can apply the given load resistance. Or, a more convenient one if we like: such as \$0\:\Omega\$. Let's do the latter (short \$A\$ to \$B\$.) We'll keep \$I\$ as the first loop's current (left side) but now we need another loop current, which I will call \$I_\text{X}\$ for the right side loop that goes through this new \$0\:\Omega\$ resistor at the output.
$$\begin{align*}
0\:\text{V}+48\:\text{V}-I\cdot R_1 - I\cdot R_2 - \left(I-I_\text{X}\right)\cdot R_3&amp;= 12\:\text{V}\\\\
0\:\text{V}+12\:\text{V}-\left(I_\text{X}-I\right)\cdot R_3-I_\text{X}\cdot R_4&amp;=0\:\text{V}
\end{align*}$$
We don't really care about \$I\$ but the above solves out as \$I_\text{X}=3.84\:\text{A}\$. So \$V_\text{A}\$ went from \$19.2\:\text{V}\$ to \$0\:\text{V}\$ while the current from \$A\$ to \$B\$ went from \$0\:\text{A}\$ to \$3.84\:\text{A}\$. From this, we find the equivalent source resistance from the perspective of \$A\$ as being \$\frac{19.2\:\text{V}}{3.84\:\text{A}}=5\:\Omega\$.
Again, same results from varying methods.