(A) The bulb,as a point source,radiates light in all directions uniformly. At a distance of $r = 3\; m$,the surface area of the surrounding sphere is $A = 4\pi r^2 = 4\pi(3)^2 = 113\; m^2$.
The power radiated by the bulb is $P = 100\; W \times 2.5\% = 2.5\; W$.
The intensity $I$ at this distance is $I = \frac{P}{A} = \frac{2.5\; W}{113\; m^2} \approx 0.022\; W/m^2$.
The average energy density is related to the electric field by $I = \varepsilon_0 E_{rms}^2 c$. Thus,$E_{rms} = \sqrt{\frac{I}{\varepsilon_0 c}}$.
Substituting the values: $E_{rms} = \sqrt{\frac{0.022}{(8.85 \times 10^{-12})(3 \times 10^8)}} \approx 2.87\; V/m \approx 2.9\; V/m$.
The peak electric field $E_0 = \sqrt{2} E_{rms} = \sqrt{2} \times 2.9 \approx 4.1\; V/m$.
The root mean square magnetic field is $B_{rms} = \frac{E_{rms}}{c} = \frac{2.9}{3 \times 10^8} \approx 9.7 \times 10^{-9}\; T$.
The peak magnetic field $B_0 = \sqrt{2} B_{rms} = \sqrt{2} \times 9.7 \times 10^{-9} \approx 1.37 \times 10^{-8}\; T$.