Do brick walls effectively protect against ionizing radiation from a nuclear explosion? Of course it depends on the distance and the amount of radiation, so let's say its about 10,000 Rad. Now, can wall made of brick protect someone behind it from such radiation ? 
 A: Chris addressed the basic matter of gammas, but neutrons are a special case and one of some interest with thermonuclear weapons.  Because they do not interact strongly with electric charge, neutrons scatter and lose energy almost entirely by contact scattering off of nuclei.
They leave the neighborhood of the explosion with considerable kinetic energy. And lose it in a series of (mostly elastic) interaction with the nuclei of the air, ground and any infrastructure that remains. Where they do interact inelastically they usually generate one of more energetic gamma rays.
With their ten minute halflife they are able to cover quite a lot of ground even after they thermalize (lose their initial kinetic energy, and come into thermal equilibrium with the surrounding material) in part because they ride along with any bulk movements of air masses.
Eventually they either decay ($n \to p + e^- + \bar\nu$) which is relatively little beta radiation  or capture (mostly on hydrogen--2.2 MeV of gamma--or carbon--4.95 MeV).
Now here's the nasty part: both the initial high energy penetration and all that Brownian motion as the thermalized neutrons bounce around in equilibrium is largely unshielded, the neutrons can get in almost anywhere. So those capture gammas can be generated inside your shielding or even inside your body. Ouch.
To shield against energetic neutrons you want as many nuclei as possible. Wax, water and plastics are common choices. To shield against thermal neutrons some dense material doped with strong neutron absorption agent. Borated plastics are good for both and are common in reactor shielding. Especially borated PVC as the chlorine is almost as helpful as the boron. Given the recent drop in the cost of gadolinium I expect to see an increasing number of uses of it in neutron shielding application.
A: I'll focus on gamma rays (and high-energy X-rays), since alpha and beta particles are easily stopped. Neutrons are slightly more complicated, since they tend to care about which nuclei they are encountering.

General theory
The answer is any material can block gamma rays so long there is enough of it. In fact, the amount of stuff is basically the only criterion there is.
The thing is, every time a gamma-ray photon interacts with a nucleus or an electron, either it is completely lost or it loses some energy. And gamma rays will interact with anything along their path. They don't care what kind of material they are in - since their wavelengths are so short, they don't "see" the molecular and large-scale structure. The fraction of gamma ray energy that passes through some matter, then, only cares about the column density of the matter. There is an exponential dependence - if your wall is letting half the radiation through, doubling its thickness will result in only a quarter getting through.
For example, if you had a brick wall that was thick enough such that a $1\ \mathrm{cm}^2$ core through it had a mass of $1\ \mathrm{kg}$ (roughly $4\ \mathrm{m}$ thick, for some "typical" brick), then it would be equivalent, in terms of gamma-ray stopping power, to


*

*A $1\ \mathrm{m}$ thick wall of lead,

*A $10\ \mathrm{m}$ thick wall of water, or

*The atmosphere (i.e., the same attenuation as between a point at sea level and a point in orbit directly above that one).



How much the material matters
My claim of independence of material is, needless to say, a little bit of a simplification. Perhaps the biggest oversight is the variation of electron-to-mass ratio (equivalently, $Z/A$ for the atoms involved). To the extent that this ratio is constant (and it pretty much is for everything from helium to about nickel), the number of particles that can stop gamma rays is proportional to mass column density, assuming the interactions are dominated by Compton scattering. If the energy of the gamma rays is in the pair-production regime, this might depend more on the number of nuclei rather than number of electrons.
Let's put these statements in graphical form. The figure below, from Wikimedia, shows how much high-energy photons are attenuated (black line) when passing through iron, as a function of how energetic those photons are. The colored lines indicated the different things that can happen to the photon, and they add up to the black line.

Now let's take a look at something closer to brick. The US National Institute of Standards and Technology (NIST) has compiled lots of data on this sort of thing here[1]. While I could not find data for "brick," they do have it for concrete here, reproduced below.

We are interested in the solid line, which shows energy lost by the radiation. The dashed line shows energy taken in by the material, which can be slightly less if some energy is lost, say, to escaping neutrinos. (The dashed line is more important in biological materials, where you care how much energy is deposited more than how much is lost from the blast.) Note that it is basically the same as for iron. In fact, to prove that these numbers don't change much as a claimed above, below are the figures for water (found here) and air (found here).



Plugging in numbers
So how exactly are these charts used? Let's take the concrete chart, and let's take the density of brick to be $\rho = 2\ \mathrm{g}/\mathrm{cm}^3$. Multiplying the ordinate in that chart by this density gives the attenuation coefficient. Suppose we are dealing with photons of energy $10^{-2}\ \mathrm{MeV} = 10\ \mathrm{keV}$. Then the attenuation coefficient is
$$ \mu_{10\ \mathrm{keV}} = (10^2\ \mathrm{cm^2}/\mathrm{g}) (2\ \mathrm{g}/\mathrm{cm}^3) = 2\times10^2\ \mathrm{cm}^{-1}. $$
If instead we are dealing with higher-energy photons, around $10\ \mathrm{MeV}$ and above, then the attenuation coefficient is
$$ \mu_{10\ \mathrm{MeV}} = (2\times10^{-2}\ \mathrm{cm^2}/\mathrm{g}) (2\ \mathrm{g}/\mathrm{cm}^3) = 4\times10^{-2}\ \mathrm{cm}^{-1}. $$
To be on the safe side, let's use the latter value, which corresponds to harder-to-stop photons, since I cannot find any data on the spectrum of ionizing radiation.
Now a wall with thickness $x$ will allow a fraction $\mathrm{e}^{-\mu x}$ of radiation through. With $\mu = 4\times10^{-2}\ \mathrm{cm}^{-1}$ and $x = 15\ \mathrm{cm}$, this comes out to be $55\%$. That is, the wall doesn't even stop half the high-energy gamma radiation. (If you use the other vale of $\mu$, you will find an answer so close to $0$ that most calculators cannot distinguish the two; low-energy radiation is completely stopped by this wall.)

Brief digression on units
There are four stages of calculating effects of radiation, as explained nicely by the US Nuclear Regulatory Commission (NRC) on this page. Each one has an SI unit and another commonly used unit to measure it, which is why measuring radiation can be so confusing. A source of radiation has its radioactivity measured in becquerels ($\mathrm{Bq}$) or curies ($\mathrm{Ci}$). The amount of radiation traveling away from the source is measured in coulombs per kilogram ($\mathrm{C}/\mathrm{kg}$) or roentgens ($\mathrm{R}$). The amount of energy actually absorbed by your body (think those dashed lines in the above figures) is measured in grays ($\mathrm{Gy}$) or rads ($\mathrm{rad}$). Finally, that last pair can be weighted by some factor to account for different types of radiation being more harmful because they induce secondary reactions in your body. The result is sieverts ($\mathrm{Sv}$) and rems ($\mathrm{rem}$). Fortunately that weight is generally taken to be $1$ for gamma rays. Quoting the NRC,

"For practical purposes, $1\ \mathrm{R}$ (exposure) = $1\ \mathrm{rad}$ (absorbed dose) = $1\ \mathrm{rem}$ or $1000\ \mathrm{mrem}$ (dose equivalent)."


Putting everything together
The $15\ \mathrm{cm}$ wall will reduce the amount of radiation reaching you by $45\%$ in the worst case scenario (all radiation is high-energy), which means the energy absorbed by your body will also be reduced by $45\%$, as will the equivalent dosage. Thus you would receive about $5500\ \mathrm{rem} = 55\ \mathrm{Sv}$. The rem article tells us,

"Doses greater than $100\ \mathrm{rem}$ received over a short time period are likely to cause acute radiation syndrome (ARS), possibly leading to death within weeks."

To reduce this to the typical amount received in a CT scan, about $1\ \mathrm{rem}$ according to the NRC, you would need a total wall thickness of
$$ x = -\frac{1}{\mu} \log\left(\frac{1\ \mathrm{rad}}{10^4\ \mathrm{rad}}\right) \approx 230\ \mathrm{cm}. $$

 [1] J. H. Hubbell and S. M. Seltzer. Tables of X-Ray Mass Attenuation Coefficients and Mass Energy-Absorption Coefficients from $1\ \mathrm{keV}$ to $20\ \mathrm{MeV}$ for Elements $Z = 1$ to $92$ and $48$ Additional Substances of Dosimetric Interest. 
