Answer
The other resistor is identical, and also has a value of $4.8\;k\Omega$.
Work Step by Step
When the resistors are in series, the power used is one-fourth the power used when they are in parallel. The applied voltage is the same in each case.
$$\frac{V^2}{R_{series}}=\frac{1}{4}\frac{V^2}{R_{parallel}}$$
$$R_{series}=4R_{parallel}$$
$$R_1+R_2=4\frac{R_1R_2}{R_1+R_2}$$
$$(R_1+R_2)^2=4R_1R_2$$
$$R_1=R_2$$
The other resistor is identical, and also has a value of $4.8\;k\Omega$.