For the network shown, the equivalent Thevenin voltage and Thevenin impedance as seen across terminals ‘ab’ is

The correct option is 65 V in series with 15 Ω.

To find the equivalent Thevenin voltage ($V_{\mathrm{th}}$) and Thevenin impedance/resistance ($R_{\mathrm{th}}$) across terminals 'ab', we proceed with the following steps:

Step 1: Finding the Thevenin Voltage ($V_{\mathrm{th}}$)

The Thevenin voltage is the open-circuit voltage across terminals 'ab' ($V_{\mathrm{oc}}=V_{\mathrm{ab}}$).
When terminals 'ab' are open-circuited, no current flows through the $2\Omega$ resistor and the dependent voltage source.
Therefore, the entire current from the $5\text{A}$ independent current source flows through the parallel $10\Omega$ resistor:

$i_1=5\text{A}$

Let the node above the $10\Omega$ resistor be node $c$ and the bottom terminal be the reference node $b$ ($V_b=0\text{V}$).
The voltage at node $c$ is:

$V_c=i_1\times 10=5\times 10=50\text{V}$

The dependent voltage source has a value of $3i_1$ with the positive terminal on the right and the negative terminal on the left (connected to node $c$).
Since no current flows through the $2\Omega$ resistor, the voltage at terminal $a$ is the same as the voltage at the positive terminal of the dependent source:

$V_{\mathrm{th}}=V_{\mathrm{ab}}=V_c+3i_1$

Substituting the value of $i_1=5\text{A}$ and $V_c=50\text{V}$:

$V_{\mathrm{th}}=50+3\left(5\right)=50+15=65\text{V}$

Step 2: Finding the Thevenin Resistance ($R_{\mathrm{th}}$)

To determine the Thevenin resistance, we deactivate the independent $5\text{A}$ current source (replacing it with an open circuit) and apply a test current source $I_x$ entering terminal $a$ and leaving terminal $b$.
Let $V_x$ be the voltage across terminals 'ab' due to the test current.
Since the independent current source is open-circuited, the current $I_x$ flows directly through the $2\Omega$ resistor, the dependent source, and downwards through the $10\Omega$ resistor.
This gives the relation:

$i_1=I_x$

Applying Kirchhoff's Voltage Law (KVL) along the path from terminal $a$ to terminal $b$:

$V_x=(I_x\times 2)+3i_1+(i_1\times 10)$

Substituting $i_1=I_x$:

$V_x=2I_x+3I_x+10I_x=15I_x$

The equivalent Thevenin resistance is:

$R_{\mathrm{th}}=\frac{V_x}{I_x}=15\Omega$

Conclusion

The equivalent Thevenin network consists of a voltage source of $65\text{V}$ in series with an impedance/resistance of $15\Omega$.