The intensity of the light from a bulb incident on a surface is 0 ....

The intensity is defined as energy per unit time per unit area. Therefore, $I = \dfrac{dE}{Adt}$.

Energy density is energy per unit volume. Therefore, ${U}_{d} = \dfrac{dE}{Adx}$.

Dividing $\dfrac{I}{{U}_{d}} = \dfrac{dx}{dt} = c$

$⇒ {U}_{d} = \dfrac{I}{c} = \dfrac{0 . 22}{3 \times {10}^{8}} = \dfrac{22}{3} \times {10}^{- 10}J{m}^{- 3}$

Magnetic energy density will be half of total energy density, Therefore, ${U}_{b} = \dfrac{11}{3} \times {10}^{- 10}J{m}^{- 3}$.

Now energy density in magnetic field is given by, ${U}_{b} = \dfrac{{B}^{2}}{2{\mu}_{0}}$. Therefore,

${B}_{rms}^{2}2{\mu}_{0} = \dfrac{11}{3} \times {10}^{- 10}J{m}^{- 3}$ where ${B}_{rms}^{2} = \left(\dfrac{{B}_{0}}{\sqrt{2}}\right) = {B}_{0}^{2}2$

$⇒ {B}_{0}^{2}4{\mu}_{0} = \dfrac{11}{3} \times {10}^{- 10} ⇒ {B}_{0} = \sqrt{4 \times \left(4\pi \times {10}^{- 7}\right) \times \dfrac{11}{3} \times {10}^{- 10}} ⇒ {B}_{0} = 4 . 29 \times {10}^{- 8}T ≃ 43 \times {10}^{- 9}T$