Barry gives the correct answer. What follows is the mathematical justification.
The power delivered to the load resistor, for a given Thevenin equivalent circuit is:
\$P_L = V_L \cdot I_L = V_{th}\dfrac{R_L}{R_{th}+ R_L} \cdot \dfrac{V_{th}}{R_{th}+ R_L} = V^2_{th} \dfrac{R_L}{(R_{th}+ R_L)^2}\$
If we fix \$R_{th}\$ and ask which value for \$R_L\$ gives maximum \$P_L\$, the value is given by \$R_L = R_{th}\$ and the resulting power delivered to the load is:
\$P_{L,max} = \dfrac{V^2_{th}}{4R_{th}} \$
However, if we fix \$R_L \$, and ask which value for \$R_{th}\$ gives maximum \$P_L\$, the answer is, by inspection, \$R_{th} = 0\$.
Thus, the answer is \$R = 500 \Omega \$ so that \$R_{th} = 0\$
For maximum power transfer you are expected to not only know how to set up equivalent circuits, but also how to find maximums and minimums by setting a derivative to zero.
The first step, of course, is to develop the equivalent minimum circuit:
simulate this circuit – Schematic created using CircuitLab
You should have been able to follow the above steps. If not, then you have some serious study to do and you won't yet be able to solve these problems until you can perform the above steps on your own.
Now that you have the final circuit, you can work out the following equation:
$$\begin{align*}
I &= \frac{V}{R+R_3} \\
\\
P_3 &= I^2 R_3 = \frac{V^2 R_3}{\left(R+R_3\right)^2}
\end{align*}$$
Now, in the above equation for \$P_3\$, \$V\$ and \$R\$ are known constants. \$R_3\$ is the variable whose value you want to find in order to maximize \$P_3\$. This is the point where you are supposed to know from calculus courses (or reading) that you take the derivative, set that new expression to zero as an equation (to find where the slope is zero), and then solve. This will give you either a minimum or a maximum.
$$\begin{align*}
\frac{\textrm{d}P_3}{\textrm{d} R_3}&=V^2\frac{R-R_3}{\left(R+R_3\right)^3} \\
\\
\textrm{so,} \\
\\
V^2\frac{R-R_3}{\left(R+R_3\right)^3}&= 0
\end{align*}$$
I'll leave the last step of solving that answer to you. But that's all that is left.
Best Answer
This is super simple; It's 12kOhm.
Why?
The 18kOhm resistor is in parallel with a voltage source, ie. it has NO influence on anything, throw it away..
The 6kOhm resistor is in series with a current source ie. it has NO influence on anything, throw it away..
What are you left with?
And so what is the output impedance of your circuit???
To get maximum power transfer you just need the load to have the same value as the output impedance of the circuit.
This should be easy enough for you to solve now. (He did solve it)