Thevenin Voltage

The circuit diagram can be redrawn as above for easy reference.gn

The voltage across $CG$ is as below:

$\begin{aligned} V_{CG} & = \frac {6||(2+4)}{3+6||(2+4)} V_{AG} \\ & = \frac {6||6}{3+6||6} V_{AG} \\ & = \frac {3}{3+3} \times 24 \\ & = 12 \text{ V} \end{aligned}$

Now, we have:

$\begin{aligned} V_{Th} & = V_{AB} \\ & = V_{AC} + V_{CB} \\ & = V_{AG} - V_{CG} + \frac 4{2+4}V_{CG} \\ & = 24 - 12 + \frac 26 \times 12 \\ & = \boxed{16} \text{ V} \end{aligned}$