Solution:

The resistance of the bulb is given by \(R = \frac{V^2}{P} = \frac{200^2}{100} = 400 \, \Omega\). When connected to a \(160 \text{ V}\) supply, the power consumed is \(P' = \frac{V'^2}{R} = \frac{160^2}{400} = 64 \text{ W}\).