The peak value of electric field produced by the radiation coming from a $100watt$ bulb at a distance of $3m$ is ( assume that the efficiency of the bulb is 2.5% and it is a point source)

Useful intensity,  $I=\frac{power}{area}=\frac{100\times \frac{2.5}{100}}{4\pi {\left(3\right)}^2}$
 $\Rightarrow I=\frac{2.5}{36\pi }W{m}^{-2}$
Half of this intensity(I) belongs to electric field and half of that to magnetic field
  $$\begin{gathered} \frac{I}{2}=\frac{1}{4}{\in }_0{E}_0^{2}C \\ \Rightarrow E_0=\sqrt{\frac{2I}{{\in }_{0}C}} \end{gathered}$$
 $\Rightarrow E_0=\sqrt{\frac{2\times (2.5/36\pi )}{\left({\displaystyle \frac{1}{4\pi \times 9\times {10}^9}}\right)\times (3\times {10}^8)}}=4.08V{m}^{-1}$