Since both lose heat at the same rate:

Q_{Z}=Q_{X}

The temperature is 10\degree C, which is \frac12 times that inside X:

T_{Z}=\frac12T_{X}\Longrightarrow\left(T_{Z}\right)^2=\frac14\left(T_{X}\right)^2

SO if T^2 is divided by 4, the ratio \frac{A}{w} must be multiplied by 4:

\frac{A_{Z}}{w_{Z}}=4\times\frac{A_{X}}{w_{X}}=4\times\frac{4}{0.2}=80

Let's think about what the internal volume is - it's the volume of the internal cube, whose side length is 2w smaller than the outer length , since there are walls of thickness w on both sides.

Let the inner side length be l, so that the inner volume is l^3=1\,\mathrm{m^3}. That means the inner length is simply 1\,\mathrm{m} .

The outer length is then 1+2w, so the area is

A_{Z}=6\left(1+2w\right)^2

So the ratio

\frac{A_{Z}}{w_{Z}}=\frac{6\left(1+2w\right)^2}{w}=80

Solving using a G.D.C:

So the solutions are

\boxed{w=0.113,2.22\,\mathrm{m}}

Problem Bank - Function Theory

Problem Bank - Function Theory