A resistor at a temperature T has a fluctuating voltage. This is a consequence of the fluctuation dissipation theorem which you can use to calculate the spectrum of the voltage. The wikipedia article on the Fluctuation Dissipation Theorem has a section on resistor thermal noise. When measured over a bandwidth $\Delta\nu$, the average squared voltage is:
$$\langle V^2\rangle = 4Rk_BT\Delta\nu$$
where $k_B= 1.38\times 10^{-23}$J/K is Boltzmann's constant.
Suppose we have a resistor $R$ maintained at a temperature T, and hooked up to a resistor $R_0$ initially at absolute zero. The thermal noise of the warm resistor will be as given above. When you apply a voltage $V$ to a resistor $R_0$, the dissipation (in watts) will be given by $IV = V^2R_0$, so the watts applied to the resistor initially at absolute zero, over a bandwidth $\Delta\nu$ will be
$$\langle V^2\rangle = 4R_0Rk_BT\Delta\nu.$$
As that resistor warms up to temperature $T_0$, it will apply a fluctuating voltage on the warm resistor. Following the above, but with the two resistors swapped, the power applied to the warm resistor by the colder resistor at temperature $T_0$ will be:
$$\langle V^2\rangle = 4RR_0k_BT_0\Delta\nu.$$
The system will be in balance when the above two powers are equal. This happens algebraically when $T=T_0$.
A possible source of paradoxical confusion is that the above calculation was done over a limited bandwidth range. But the calculation does not depend on frequency; instead the power transmitted is simply proportional to the range of bandwidths.
For the usual physical system, we consider frequencies that run from 0 to infinity. Thus the total bandwidth is infinite. This suggests that the power flow in the above should be infinite. This paradox is avoided by noting that physical resistors have a limited bandwidth. There is always a parasitic capacitance so that the bandwidth is limited on the high side. Thus the power transfer rate depends on how ideal your resistors are.
As an example calculation, suppose that a resistor has a maximum frequency of 100 GHz $= 10^{11}$ Hz, a (room) temperature of 300K, and a resistance of 1000 ohms. Then the power transfer rate is:
$$ 4\times 1000 \times 1.38\times 10^{-23} 10^{11} \times 300 = 1.66\;\;\textrm{uWatts}$$
Given the heat capacity of the resistor, you can compute the relaxation time with which the colder resistor exponentially approaches an equal temperature.
At thermal equilibrium, a blackbody will have no net emission of energy, but that's not the same thing as no emission at all. Obviously, a blackbody will emit blackbody radiation, but at equilibrium it will absorb exactly the same amount of energy. If it emits more than it absorbs, its temperature will fall; if it absorbs more than it emits its temperature will rise. In neither case will it be at thermal equilibrium.
I suspect whoever wrote the exam was being very careless.
Best Answer
In general, it doesn't hold. It only holds for thermal radiation emission (when the radiation is due to microscopic thermal motions characterized by thermodynamic temperature, not due to other non-equilibrium processes that are not characterized by thermodynamic temperature).
Consider a fluorescent mercury gas discharge lamp. When supplied with electric energy, its phosphor layer (a chemical compound, not the element) produces much more visible radiation energy than it absorbs and ratio of these energies isn't determined by temperature of the radiating surface. So this radiating surface doesn't obey Kirchhoff's radiation law. A different way to describe this is to say that light of a discharge lamp is not due to thermal emission. Similarly for the laser light.
In some non-equilibrium cases Kirchhoff's radiation law is nevertheless obeyed. For example, if the body surface radiates just thermal radiation (such as in your example of energy supplied via heat conduction), then there is no reason why the law should not be obeyed. The surface elements do not "know" that their temperature is due to heat conduction from a hotter body and not from incoming radiation.