Answer
(a) $$1$$
(b) $$c = 2 or 4$$
Work Step by Step
(a) $$f_{ave} = {1\over 3} \int ^5 _2 (x-3)^2 dx = {1\over 3} [{1\over 3} (x-3)^3]^5 _2$$
$$={1\over 9} [2^3 - (-1)^3] = {1\over 9} (8+1) = 1$$
(b) $$f(c) = f_{ave}$$
$$(c-3)^2 = 1$$
$$c-3 = \pm1$$
$$c = 2 or 4$$