Answer
The ratio $R_1/R_2$ is 4:1.
Work Step by Step
As the problem notes, the circuit is the same as in Example 19-3. The ratio $R_1/R_2$ should be 4:1.
Both resistors, because they are in series, have the same current passing through each of them.
V=IR, so the voltage drop across $R_1$ will be 4 times that across $R_2$, which is the requirement.
A circuit with $R_1=400\Omega$ and $R_2=100\Omega$ would suffice.
