The mean intensity of radiation on the surface of the Sun is about $$10^8 W/m^2$$. The rms value of the corresponding magnetic field is closest to:

We are given that the mean intensity of solar radiation is $$I = 10^{8}\;{\rm W\,m^{-2}}$$. Our aim is to find the rms (root-mean-square) value of the accompanying magnetic field.

For an electromagnetic wave travelling in free space, the average intensity is related to the amplitudes of the fields by the well-known formula

$$I \;=\; \dfrac{c}{2\mu_{0}}\;B_{0}^{2},$$

where

$$c = 3\times 10^{8}\;{\rm m\,s^{-1}} \quad\text{and}\quad \mu_{0}=4\pi\times10^{-7}\;{\rm H\,m^{-1}}.$$

Here $$B_{0}$$ is the peak (maximum) value of the magnetic field. The rms value is connected to the peak value through

$$B_{\rm rms}=\dfrac{B_{0}}{\sqrt{2}}.$$

Combining the two relations, we first solve the intensity formula for $$B_{0}$$:

$$B_{0}^{2} \;=\; \dfrac{2\mu_{0}I}{c}.$$

Taking the square root,

$$B_{0} \;=\; \sqrt{\dfrac{2\mu_{0}I}{c}}.$$

Dividing by $$\sqrt{2}$$ to convert to rms,

$$B_{\rm rms} \;=\; \dfrac{1}{\sqrt{2}}\;\sqrt{\dfrac{2\mu_{0}I}{c}} \;=\;\sqrt{\dfrac{\mu_{0}I}{c}}.$$

Now we substitute the numerical values:

$$\mu_{0}I \;=\;(4\pi\times10^{-7})\times10^{8} \;=\;4\pi\times10^{1} \;\approx\;1.256\times10^{2},$$

and therefore

$$\dfrac{\mu_{0}I}{c} \;=\;\dfrac{1.256\times10^{2}}{3\times10^{8}} \;=\;4.187\times10^{-7}.$$

Taking the square root gives

$$B_{\rm rms} \;=\;\sqrt{4.187\times10^{-7}} \;=\;2.046\times10^{-3.5} \;=\;2.046\times3.162\times10^{-4} \;\approx\;6.5\times10^{-4}\;{\rm T}.$$

This value is of the order $$10^{-4}\;{\rm T}$$ and is closest to the fourth option in the list.

Hence, the correct answer is Option D.