The rate of heat flow through a conductor is:

\(\displaystyle{H}={\left({\frac{{{k}{A}}}{{{L}}}}\right)}\triangle{T}\) where k is the conductivity of the material, A is the surface area, L is the thickness

You can write this for wool sweater and for down jacket:

\(\displaystyle{H}_{{w}}={\left({\frac{{{k}_{{w}}{A}}}{{{L}_{{w}}}}}\right)}\triangle{T}\) and \(\displaystyle{H}_{{d}}={\left({\frac{{{k}_{{d}}{A}}}{{{L}_{{d}}}}}\right)}\triangle{T}\)

And take the ratio of these equations:

\(\displaystyle{\left({\frac{{{H}_{{w}}}}{{{H}_{{d}}}}}\right)}={\frac{{{\frac{{{k}_{{w}}}}{{{L}_{{w}}}}}}}{{{\frac{{{k}_{{d}}}}{{{L}_{{d}}}}}}}}={\frac{{{k}_{{w}}{L}_{{d}}}}{{{k}_{{d}}{L}_{{w}}}}}\)

my book doesnt have the k values for wool ordown, but yours must have them if its asking this question. Popthem in and calc the answer. Notice the units drop out and you willjust have a number that represents how much more heat is lost persecond through the sweater than through the down.

\(\displaystyle{H}={\left({\frac{{{k}{A}}}{{{L}}}}\right)}\triangle{T}\) where k is the conductivity of the material, A is the surface area, L is the thickness

You can write this for wool sweater and for down jacket:

\(\displaystyle{H}_{{w}}={\left({\frac{{{k}_{{w}}{A}}}{{{L}_{{w}}}}}\right)}\triangle{T}\) and \(\displaystyle{H}_{{d}}={\left({\frac{{{k}_{{d}}{A}}}{{{L}_{{d}}}}}\right)}\triangle{T}\)

And take the ratio of these equations:

\(\displaystyle{\left({\frac{{{H}_{{w}}}}{{{H}_{{d}}}}}\right)}={\frac{{{\frac{{{k}_{{w}}}}{{{L}_{{w}}}}}}}{{{\frac{{{k}_{{d}}}}{{{L}_{{d}}}}}}}}={\frac{{{k}_{{w}}{L}_{{d}}}}{{{k}_{{d}}{L}_{{w}}}}}\)

my book doesnt have the k values for wool ordown, but yours must have them if its asking this question. Popthem in and calc the answer. Notice the units drop out and you willjust have a number that represents how much more heat is lost persecond through the sweater than through the down.