mean value over interval

$\bar{x} = \dfrac{1}{3-0} \displaystyle \int \limits_0^3~2x^2-3~dx = \\ \dfrac 1 3 \left(\left. \dfrac 2 3 x^3 - 3x \right |_0^3\right) = \\ \dfrac 1 3 \left(18-9\right) = \\ \dfrac 9 3 = 3$

I suspect you omitted the factor of 1/3 used to normalize the integral

I assume you can do part (b)