How far extra low frequency electromagnetic waves can go away from Sun? Since ELF EM (3 Hz to 30 Hz) has a large wavelength (100000 km to 10000 km), i was wondering that how far from Sun these waves can go? And if radio waves that are used in mobile communications can pass through walls and buildings, how deep ELF EM can go inside the Earth?
 A: There are two points to consider. One is whether the waves are reflected from the surface (or at any interface between layers with different electrical properties) and second, how far waves can propagate once inside a partially conductive medium.
The transmission of a certain frequency into a medium is approximately proportional to its impedance. In turn, the impedance is proportional to $\sqrt{f/\sigma}$, where $\sigma$ is the conductivity of the medium. Thus low frequencies do not necessarily help if the conductivity is high. The penetration depth of the wave in a good conductor is approximately proportional to $\sqrt{1/\sigma f}$, so low frequencies and low conductivity lead to larger penetration.
The conductivity of "earth" (typical soil), might be of order 0.01 S/m. Sea water is more conductive - about 5 S/m.
Whether they act as good conductors is determined by the ratio $\sigma/2\pi \epsilon_0 f$. Taking $f=3-30$ Hz, we see that both soil and seawater are excellent conductors at these frequencies.
In those circumstances, the amplitude transmission fraction is aporoximately
$$ t \simeq \frac{2\eta}{377},$$
where 377 Ohms is the vacuum impedance and $\eta \simeq \sqrt{2\pi f \mu_0/\sigma}$ is the impedance of the medium. So the transmission fraction is much less than 1 percent for both these materials.
Once in the medium,  the signal fades exponentially, with a penetration depth (where that amplitude falls by $1/e$)  given approximately by
$$ d = \sqrt {\frac{2}{2\pi f \mu_0 \sigma}}$$,
which is 1-3 km for soil and is 40-130 m for seawater.
Thus my answer would be, most of the waves are reflected, but for those that penetrate, the depth reached is a few km if not in the ocean and a few hundred metres in the ocean (and indeed ELF is used for communicating with submerged submarines).
A second part to your question concerns propagation of waves through space.
Here, the relevant point is whether the wave frequency is above or below the plasma frequency
$$\nu_p = \left( \frac{e^2 n_e}{4\pi^2 \epsilon_0 m_e}\right)^{1/2} = 9000 \left(\frac{n_e}{{\rm cm}^{-3}}\right)^{1/2}\ {\rm Hz},$$
where $n_e$ is the electron number density and $m_e$ is the electron mass. If waves have a frequency below the plasma frequency then they will be reflected.
The value of $n_e$ varies considerably from place to place. In the solar corona, typical values are $n_e \sim 10^{5}$ to $10^{6}$ cm$^{-3}$. This means that ELF waves generate by the Sun will not propagate through the coronal plasma. In addition, the electron density in the ionosphere of the Earth is typically of the same order of magnitude. So ELF waves from outer space will not penetrate the Earth's ionosphere.
Even should the ELF waves be produced near the top of the corona, they will not propagate through the solar wind. The electron density falls with distance from the Sun, but is still $n_e \sim 100$ cm$^{-3}$ at the orbit of the Earth, giving a plasma frequency of 90 kHz.
A: I just want to comment on one point and this probably will not answer your question completely. LF frequencies correspond to very large wavelengths such that LF waves have wavelengths that match the geometrical extent of vertical refractive distribution of the atmosphere. This provides the travelling LF wave to couple to the surface (does not travel straight out to open space) and travel far distances with almost no loss. They are observed to travel round and round around the world. So, LF wave propagation is known to have small propagation loss for this reason. 
However, you are talking about LF waves originating from the Sun at open space. I believe it is different and you should not assume the waveguide coupling advantage of the atmosphere to be valid in space. I hope this helps.
