Explanation

(d) The light from bulb spread out uniformly in all directions.
For a 100 W bulb, intensity at a distance of 3 m is
$I = \frac{Power}{Area} = \frac{100}{4\pi(3)^{2}}$
As$\ I = \varepsilon_{0}cE_{rms}^{2} \Rightarrow E_{rms}^{2} = \frac{I}{\varepsilon_{0}c}\ \ \ \ \ $ …(i)
For a 400 W bulb, intensity at the same point is
$I^{'} = \frac{400}{4\pi(3)^{2}} \Rightarrow E_{rms}^{'2} = \frac{I^{'}}{\varepsilon_{0}c}$
$\frac{E_{rms}^{'2}}{E_{rms}^{'2}} = \frac{I^{'}}{I} = \frac{400}{4\pi(3)^{2}} \times \frac{4\pi(3)^{2}}{100}$
$E_{rms}^{'2} = E_{rms}^{2} \times 4 = (2.9)^{2} \times 4\ \lbrack\because E_{rms} = 2.9Vm^{- 1}$(Given)]
or $E_{rms}^{'} = 2.9 \times 2Vm^{- 1} = 5.8Vm^{- 1}$