Great Explanation

Now look better, you forgot the tag

Step 1: Understanding the question
Let A, B and O be the centers of the circles and radius of the smaller circle be x
As the diagonals of the square intersect at right angle, triangle AOB is a right angled at O

AB = 4+4 = 8
AO = (x+4)
BO = (x+4)

Applying Pythagoras
$AB^2 = AO^2 + BO^2$
64 = 2* $(x+4)^2$
32 = $(x+4)^2$
x + 4 = 4$\sqrt{2 }$
x = 4$\sqrt{2}$ - 4

Hence diameter of the smaller circle = 8$\sqrt{2}$ - 8
C is correct

Simple way is just joining all the centres of the large circle to form a square of side 8 units.
Joining the diagonal of the square that cuts the smaller circle at its diameter.
so diagonal of a square = sqrroot 2 * a = 8 * sqr root 2

Now diameter of smaller circle = diagonal - (radius of 2 bigger circle) = 8* sqr root 2 - (4+4) = 8 * sqr root 2 - 8