About 5% of the power of a 100 W light bulb is converted to visible radiation.

Power rating of bulb, P = 100 W

It is given that about 5% of its power is converted into visible radiation.

$\therefore$ Power of visible radiation,

$P^{\prime}=\frac{5}{100} \times 100=5 \mathrm{~W}$

Hence, the power of visible radiation is 5W.

(a) Distance of a point from the bulb, d = 1 m

Hence, intensity of radiation at that point is given as:

$I=\frac{P^{\prime}}{4 \pi d^{2}}$

$=\frac{5}{4 \pi(1)^{2}}=0.398 \mathrm{~W} / \mathrm{m}^{2}$

(b) Distance of a point from the bulb, d1 = 10 m

Hence, intensity of radiation at that point is given as:

$I=\frac{P^{\prime}}{4 \pi\left(d_{1}\right)^{2}}$

$=\frac{5}{4 \pi(10)^{2}}=0.00398 \mathrm{~W} / \mathrm{m}^{2}$