Convection and radiation heat transfer coefficients

I have a problem where we have given conditions:

• Inner temperature is: 20 celcius
• Outer temperature is: -5 celcius

Between these temperatures we have a wall, made of:

• 50mm of mineral wool with thermal conductivity of $lambda=0.04frac{ W}{mK}$
• and 270mm of tile with thermal conductivity of $lambda=0.6frac{ W}{mK}$

Tile is on the inner side of the wall and mineral wool is on the outer side.

Now I need to figure out the heat transfer coefficient of convection and radiation.

I have tried to take Gr and Pr from VDI Heat Atlas table for properties of dry air. With that I got $R_a = Gr*Pr = 127 000$ and Nu = 0.092 *127 0001/3 = 4.6.

Then I calculated the coeffiecient of convection (a):

$$a = frac{Nu * λt}{ Lt},qquad jossa quad lambda t = frac{4.6 * 0.6 W/m·K}{ 0.27 m} = 10.22… approx 10.2$$

and for coefficient of radiation (b):

$$
b = epsilon sigma(T_{p2} – T_{y2})(T_{p} – T_{y})$$

$$b = 0.93 cdot 5.67cdot 10^{-8} frac{ W}{m^2K^4} cdot (287.952 + 293.152) cdot ( 287.95 + 293.15) $$
$$ b= 5.173… approx 5.17 $$

I took the value from emissivity from a website that specified emissivities for different types of tiles, not sure if it’s right.

Now, shouldn’t thermal resistance for inner surface be 0.13 and reciprocal of total heat transfer coefficient. With my values, it isn’t:

1/(10.2+5.17) = 0.065 != 0.13

What is wrong with the calculations?