A rectangular box is inscribed in a sphere of radius r cm. Surface area of the box is 384 cm$$^2$$ and the sum of lengths of its 12 edges is 112 cm. The value of r is

Let l,b,h be the dimensions of the rectangular box. 

Sum of 12 edges is 112cm, which means 112 = 4l+4B+4h

Hence, l+b+h= 28

Now, the surface area of the box:

384= 2lb+2bh+2lh

This means, lb+bh+lh= 192

Also: due to the rectangular box being inscribed in the sphere, the diagonal of the box will be same as the diameter of the sphere.

Hence, $$\ \ \sqrt{\ l^2+b^2+h^2}$$=$$\ \sqrt{\ (l+b+h)^2-2\left(lh+bh+lh\right)}$$= 2r

2r= 20

r=10 cm