#### Answer

$n \geq 26$

#### Work Step by Step

Consider the error bound for the Trapezoidal Rule as follows:
$|E_T| \leq \dfrac{M(b-a)^3}{12n^2}$
The maximum value of $|f^{2}(x)|$ on $[0,1]$, $M=8$
Now, $|E_T| \leq \dfrac{8(1-0)^3}{12n^2}=\dfrac{2}{3n^4}$
We will choose $n$ such that $\dfrac{2}{3n^2} \leq 10^{-3}$
So, $n \geq 25.8 \approx 26$
and $n \geq 26$