Answer
a) $ 0.067 \;\rm V/m$
b) $7.076\rm\; km$
Work Step by Step
a) We know that the average energy density is given by
$$u_{avg}=\frac{1}{2}\varepsilon_0E^2$$
We also know that $E_{\rm rms}=\dfrac{E}{\sqrt2}$; thus $E=\sqrt2 E_{\rm rms}$.
$$u_{avg}=\frac{1}{2}\varepsilon_0(\sqrt2 E_{\rm rms})^2=\varepsilon_0 E_{\rm rms} ^2$$
Solving for $E_{\rm rms}$;
$$ E_{\rm rms} =\sqrt{\dfrac{u_{avg}}{ \varepsilon_0} }$$
Plugging the known;
$$ E_{\rm rms} =\sqrt{\dfrac{ 4\times10^{-14}}{ 8.85\times10^{-12}} }=\color{red}{\bf 0.067}\;\rm V/m$$
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b)
We can use the average intensity as a comparable value.
We know that the average density of an electromagnetic wave is given by
$$I_{avg}=\dfrac{P_{avg}}{A}=\dfrac{P_{avg}}{4\pi r^2}$$
We also know that the average density of an electromagnetic wave is given by $I_{avg}=\frac{1}{2}\varepsilon_0 c E_0^2$.
Thus,
$$ \frac{1}{2}\varepsilon_0 c E_0^2=\dfrac{P_{avg}}{4\pi r^2}$$
Recall that $E=\sqrt2 E_{\rm rms}$, as we found above.
$$ \frac{1}{2}\varepsilon_0 c (\sqrt 2E_{\rm rms})^2=\dfrac{P_{avg}}{4\pi r^2}$$
$$ \varepsilon_0 c E_{\rm rms}^2=\dfrac{P_{avg}}{4\pi r^2}$$
Solving for $r$;
$$ r =\sqrt{\dfrac{P_{avg}}{4\pi \varepsilon_0 c E_{\rm rms}^2}}$$
Plugging the known;
$$ r =\sqrt{\dfrac{7500}{4\pi\cdot 8.85\times10^{-12}\cdot 3.0\times10^8\cdot 0.067^2}}=7076\;\rm m$$
$$ r =\color{red}{\bf 7.076}\;\rm km$$
