When connected across a 220 V power supply, an electric bulb...

The resistance of the electric bulb is given by $R = \dfrac{V^2}{P}$

Substituting the given values, $R = \dfrac{(220)^2}{60} \, \Omega$

Assuming the resistance of the bulb remains constant, the power consumed at a new voltage $V'$ is $P' = \dfrac{(V')^2}{R}$

When the supply voltage drops to $180 \text{ V}$:

$P_1 = \dfrac{(180)^2}{\dfrac{(220)^2}{60}} = \left(\dfrac{180}{220}\right)^2 \times 60 = \dfrac{81}{121} \times 60 \approx 40.16 \text{ W}$

When the supply voltage is raised to $240 \text{ V}$:

$P_2 = \dfrac{(240)^2}{\dfrac{(220)^2}{60}} = \left(\dfrac{240}{220}\right)^2 \times 60 = \dfrac{144}{121} \times 60 \approx 71.40 \text{ W}$

The power consumed will be approximately $40 \text{ W}$ and $71 \text{ W}$.