This is an excellent question, but not readily answered, as discussed in Floris's Answer.

Here is a stab at how to get an estimate of efficiency. It is the method which I believe is reasonably accurate: the actual values will need to be refined by experimental measurement: I am not too confident of the actual numbers that fall out owing to the calculation's high sensitivity to the variables used (in particular the fourth power influence of the flame temperature).

For an open hearth, it would seem reasonable to assume that most of the heat transfer to the room is through radiation; this will make you feel warm and will also warm things in the room up which then raise the temperature through conduction and convection. This is because, as discussed in Floris's Answer, the fire draws in a great deal of air. The chimney and hearth shape are chosen so that the fire's heat sets up a considerable draught through the fire and up the chimney to keep the fire burning lustily and this would limit convection / conduction severely.

From here, I get a flame temperature for burning wood of about $1300{\rm K}$ (i.e. $1027{\rm ^o C}$). From here I get a $\Delta H$ for burning wood of about $15{\rm MJ \,kg^{-1}}$. Assume that we can burn $50{\rm kg}$ of wood for one hour to give a flame area of $\frac{1}{2}{\rm m^2}$.

Then, the total heat output ($\Delta H$) per second is $50\times 15\times 10^6/3600\approx 208{\rm kW}$.

The total radiant power, from the Stefan Boltzmann law is of the order of $\frac{1}{2}\times\sigma\,T^4$ (with a $\frac{1}{2}{\rm m^2}$ flame area as seen from the room), i.e. $\frac{1}{2}\times 5.7\times 10^{-8}\times 1300^4\approx 81{\rm kW}$.

So our efficiency here is estimated to be in the tens of percent range. I don't believe it is quite as high as the $30\%$ implied here. I expected the answer to be minuscule - of the order of $1\%$. I am skeptical of this exact answer, but I believe you could get a good estimate from its method using experiments to measure wood burning rate (you'd measure this over many hours to get a good estimate of the rate), a pyrometer to probe the flame's temperature and temperature uniformity and perhaps some ad hoc image processing of a video of the fire to estimate the radiant area. You could even do some crude pyrometry with a colour video, using a laboratory pyrometer to help you calibrate your measurement. Grounded on this calculation, I would estimate between $10\%$ and $30\%$ efficiency. As I said, this is a great deal better than I thought. However, you would need to experiment to refine this method: note the very high sensitivity of the calculation to the variables used - particularly the flame temperature (which has a fourth power influence).

For a closed wood burning stove, you can see the situation would be much more complicated. The stove's housing is raised to some temperature by the reverberating furnace within: several hundreds of degrees celsius, which then heats the room by convection, conduction as well as radiation. I daresay there are heating furnace efficiency specifications and standards which will let you estimate the temperature of the stove and its power output.

This is an excellent question, but not readily answered, as discussed in Floris's Answer.

Here is a stab at how to get an estimate of efficiency. It is the method which I believe is reasonably accurate: the actual values will need to be refined by experimental measurement: I am not too confident of the actual numbers that fall out owing to the calculation's high sensitivity to the variables used (in particular the fourth power influence of the flame temperature).

For an open hearth, it would seem reasonable to assume that most of the heat transfer to the room is through radiation; this will make you feel warm and will also warm things in the room up which then raise the temperature through conduction and convection. This is because, as discussed in Floris's Answer, the fire draws in a great deal of air. The chimney and hearth shape are chosen so that the fire's heat sets up a considerable draught through the fire and up the chimney to keep the fire burning lustily and this would limit convection / conduction severely.

From here, I get a flame temperature for burning wood of about $1300{\rm K}$ (i.e. $1027{\rm ^o C}$). From here I get a $\Delta H$ for burning wood of about $15{\rm MJ \,kg^{-1}}$. Assume that we can burn $50{\rm kg}$ of wood for one hour to give a flame area of $\frac{1}{2}{\rm m^2}$.

Then, the total heat output ($\Delta H$) per second is $50\times 15\times 10^6/3600\approx 208{\rm kW}$.

The total radiant power, from the Stefan Boltzmann law is of the order of $\frac{1}{2}\times\sigma\,T^4$ (with a $\frac{1}{2}{\rm m^2}$ flame area as seen from the room), i.e. $\frac{1}{2}\times 5.7\times 10^{-8}\times 1300^4\approx 81{\rm kW}$.

So our efficiency here is estimated to be in the tens of percent range. I don't believe it is quite as high as the $30\%$ implied here. I expected the answer to be minuscule - of the order of $1\%$. I am skeptical of this exact answer, but I believe you could get a good estimate from its method using experiments to measure wood burning rate (you'd measure this over many hours to get a good estimate of the rate), a pyrometer to probe the flame's temperature and temperature uniformity and perhaps some ad hoc image processing of a video of the fire to estimate the radiant area. You could even do some crude pyrometry with a colour video, using a laboratory pyrometer to help you calibrate your measurement. Grounded on this calculation, I would estimate between $10\%$ and $30\%$ efficiency. As I said, this is a great deal better than I thought. However, you would need to experiment to refine this method: note the very high sensitivity of the calculation to the variables used - particularly the flame temperature (which has a fourth power influence).

For a closed wood burning stove, you can see the situation would be much more complicated. The stove's housing is raised to some temperature by the reverberating furnace within: several hundreds of degrees celsius, which then heats the room by convection, conduction as well as radiation. I daresay there are heating furnace efficiency specifications and standards which will let you estimate the temperature of the stove and its power output.