Explanation

(d) $P_{1} = \frac{(220)^{2}}{R_{1}}\ and\ P_{2} = \frac{(220 \times 0.8)^{2}}{R_{2}}$
$\frac{P_{2}}{P_{1}} = \frac{(220 \times 0.8)^{2}}{(220)^{2}} \times \frac{R_{1}}{R_{2}} \Rightarrow \frac{P_{2}}{P_{1}} = (0.8)^{2} \times \frac{R_{1}}{R_{2}}$
Here, $R_{2} < R_{1}$ (because voltage decreases from $220\ V \rightarrow 220 \times 0.8\ V$
It means heat produced $\rightarrow$ decreases)
So $\frac{R_{1}}{R_{2}} > 1 \Rightarrow P_{2} > (0.8)^{2}P_{1} \Rightarrow P_{2} > (0.8)^{2} \times 100\ W$
Also$\ \frac{P_{2}}{P_{1}} = \frac{(220 \times 0.8)i_{2}}{220i_{1}}$, Since$\ i_{2} < i_{1}$ [we expect]
So $\frac{P_{2}}{P_{1}} < 0.8 \Rightarrow P_{2}(100 \times 0.8)$
Hence the actual power would be between $100 \times (0.8)^{2}W$ and
$(100 \times 0.8)\ W$