Is there a way to find this resistance?
The power transferred to the resistor is the voltage times the current through the resistor:

$$P = U \times I$$

The voltage U over \$R_L\$ is (via voltage divider):

$$U =  \frac{R_L}{2k\Omega + 4k\Omega + R_L} \times 8V$$

The current I is 

$$I = \frac{8V}{2k\Omega + 4k\Omega + R_L}$$

Putting it all together

$$P =  \frac{R_L}{2k\Omega + 4k\Omega + R_L}\times 8V \times\frac{8V}{2k\Omega + 4k\Omega + R_L}$$

With \$2k\Omega + 4k\Omega = 6k\Omega\$, we get

$$P =  \frac{R_L\times (8V)^2}{(6k\Omega + R_L)^2} $$

The maths people say that at its extreme points, the derivative of P becomes 0.
Said derivative \$P'\$ is
$$P' = 0 = \frac{(6k\Omega - R_L)\times (8V)^2}{(6k\Omega + R_L)^3} $$
Which is sufficiently equivalent to asking
$$P' = 0 = (6k\Omega - R_L)\times (8V)^2$$

There you have it
$$R_L = 6k\Omega $$

Admittedly, this could be a maximum or a minimum. The maths people come to the rescue again, saying that it's a maximum if \$P'' &lt; 0\$, which is the case

That's it.