Nearly 10 % of the power of a 110W light bulb is converted to...

Power of visible radiation is ${P}^{'} = \dfrac{10}{100} \times 110W$ $= 11W$

Change in average intensity of visible radiation is $∆ {I}_{radiation} = {I}_{{radiation}_{1}} - {I}_{{radiation}_{2}}$

$⇒ {I}_{1} - {I}_{2} = \dfrac{{P}^{'}}{4\pi{r}_{1}^{2}} - \dfrac{{P}^{'}}{4\pi{r}_{2}^{2}}$

$= \dfrac{11}{4\pi}$\left[\dfrac{1}{1} - \dfrac{1}{25}\right] = \dfrac{11}{4\pi} \times \dfrac{24}{25}$</p><p>$= \dfrac{264}{\pi} \times {10}^{- 2} = 84 \times {10}^{- 2}W{m}^{- 2}$</p><p>Hence, the value of$a = 84.