Now, $$\dfrac{P_{2}}{P_{1}}=\dfrac{\left ( 220\times 0.8 \right )^{2}}{(220)^{2}}\times \dfrac{R_{1}}{R_{2}}$$
$$\Rightarrow \dfrac{P_{2}}{P_{1}}=(0.8)^{2}\times \dfrac{R_{1}}{R_{2}}$$
Since $$V_2<V_1$$, voltage has been decreased and is directly proportional to the resistance, from Ohm's law.
Hence, $$R_{2}< R_{1}$$
$$\Rightarrow \dfrac{R_1}{R_2} > 1$$
$$\Rightarrow \dfrac{P_2}{P_1} > (0.8)^2$$
$$\Rightarrow P_2 > 100 \times (0.8)^2 \ W$$
Also, Since Power $$P = Vi$$
Hence$$\dfrac{P_{2}}{P_{1}}=\dfrac{\left ( 220\times 0.8 \right )i_2}{220i_1},$$
Since $$V_2<V_1$$, voltage has been decreased and is directly proportional to the current, from Ohm's law.
Hence, $$i_{2}< i_{1}$$
$$\Rightarrow \dfrac{i_2}{i_1} <1$$
So $$\dfrac{P_{2}}{P_{1}}< 0.8\Rightarrow P_{2}< (100\times 0.8) \ W$$
Hence the actual power would be between $$100 \times (0.8)^{2} \ W$$ and $$100\times (0.8) \ W$$