Chapter 17 - Electric Current and Resistance - Learning Path Questions and Exercises - Exercises - Page 621: 55

$\frac{4}{3}$

$P=\frac{V^{2}}{R}$ where $P$ is the power, $V$ is the voltage and $R$ is the resitance. $\implies R_{1}=\frac{(V_{1})^{2}}{P_{1}}=\frac{(120\,V)^{2}}{60\,W}=240\,\Omega$ $R_{2}=\frac{(V_{2})^{2}}{P_{2}}=\frac{(60\,V)^{2}}{20\,W}=180\,\Omega$ $\frac{R_{1}}{R_{2}}=\frac{240\,\Omega}{180\,\Omega}=\frac{4}{3}$