A resistance 'R' draws a power 'P' when connected to A.C. source....

Let the RMS voltage of the AC source be $V$.

When only resistance $R$ is connected, the power drawn is $P = \dfrac{V^2}{R}$.

When an inductor is connected in series, the impedance becomes $Z$. The RMS current in the circuit is $I = \dfrac{V}{Z}$.

The power drawn in the L-R circuit is $P_1 = I^2 R = \left(\dfrac{V}{Z}\right)^2 R = \dfrac{V^2 R}{Z^2}$.

The ratio of the powers is $\dfrac{P_1}{P} = \dfrac{\dfrac{V^2 R}{Z^2}}{\dfrac{V^2}{R}} = \dfrac{R^2}{Z^2} =$\left(\dfrac{R}{Z}\right)^2$.

Answer: \left(\dfrac{R}{Z}\right)^2