I've done this again to be more precise before posting it here and I got a slightly different result

, it's 111.61pm.

First to calculate the length of the unit cell of diamond using the density of diamond:

ρ=m/V, m=M·n

_{C}/Na, V=a

^{3}, where a is the length of the unit cell, M is the molar mass of carbon, Na is Avogadro's number and n

_{C} is the number of species in the unit cell (8 for diamond).

I get the equation ρ=n

_{C}*M/(Na·a

^{3}). From this equation a is equal to 3.5722·10

^{-8}cm.

Now, to calculate the covalent radius of carbon. The half of the smaller diagonal of the unit cell is equal to a·sqrt(2)/2, and it is one side of a triangle that 3 C atoms are making, the other two sides are C-C bonds (as shown in the attached picture). The angle between the two C-C bonds are 109°28' (because the C atom that is in the picture connected to two C atoms is in a tetrahedral hole and therefore the tetrahedron angle). I mark the C-C bond length with x, and using the law of cosines: a

^{2}/2=2x

^{2}-2x

^{2}cos109°28' I got that x=1.5469·10

^{-8}cm. r

_{C} is the covalent radius of C and x=2r

_{C}, therefore r

_{c}=77.34pm.

The same procedure is for the unit cell of SiC. Using the formula for density I got that a=4.3634·10

^{-8}cm. Using the law of cosines (C atoms are in the tetrahedral holes), x=1.8895·10

^{-8}cm. x=r

_{C}+r

_{Si}, and finally r

_{Si}=111.61pm

The C atoms in the picture aren't in the same plane. The ones in the tetrahedral holes are a little further from the observer that the other three C atoms.