Nearly 10 % of the power of a 110   W light bulb is converted to visible radiation. The change in…

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

Change in average intensity of visible radiation is  $∆I_{\mathrm{radiation}}=I_{{\mathrm{radiation}}_1}-I_{{\mathrm{radiation}}_2}$

$\Rightarrow I_1-I_2=\frac{P^{\text{'}}}{4\pi r_1^2}-\frac{P^{\text{'}}}{4\pi r_2^2}$

$=\frac{11}{4\pi }\left[\frac{1}{1}-\frac{1}{25}\right]$  $=\frac{11}{4\pi }\times \frac{24}{25}$

$=\frac{264}{\pi }\times {10}^{-2}=84\times {10}^{-2} \mathrm{W} {\mathrm{m}}^{-2}$

Hence, the value of $a=84$.