Let $X_1,X_2,..X_{200}$ be i.i.d. random variables with $E(X_i)=0$ and $V(X_i)=1$. Prove that $P[|X_1X_2+X_2X_3+...+X_{199}X_{200}|<10]$ is approximately equal to $\int_{0}^{1} \phi(x) dx$ where $\phi(.)$ is the pdf of the standard normal distribution.
I tried to write each $X_iX_j$ as $X_i ^2$ as they are i.i.d and used the fact that $E(X_i^2)=1$, but I also required to calculate $V(X_i^2)$ which is difficult to find.
Any other tricks or manipulations to solve this?
Your question seems to be about how to compute the mean and variance of the quantity $S_n=X_1X_2+X_2X_3+\cdots +X_{n}X_{n+1}$, then you will approximate it by a Gaussian random variable with the same mean and variance to get the integral. Note it is a separate (more theoretical) question to justify why the central limit theorem applies in this case - for more on the topic, see here for example.
Anyway, the first thing to notice is that $S_n$ has mean zero. Indeed, each term of the form $X_iX_{i+1}$ satisfies $\mathbb E[X_iX_{i+1}]=\mathbb EX_i\mathbb EX_{i+1}$ (using independence) and you are multiplying zero by itself to get zero.
Now we can calculate the variance as follows: $$ \textrm{Var}(S_n)=\mathbb E[S_n^2]=\mathbb E\sum_{i,j=1}^nX_iX_{i+1}X_jX_{j+1}. $$ The kinds of terms that appear are like $X_1X_2X_3X_4$, like $X_1X_2^2X_3$, or like $X_1^2X_2^2$. Of these, only the latter have non-zero expectation (using the same reasoning as before). Therefore $$ \mathbb E[S_n^2]=\mathbb E\sum_{i=1}^nX_i^2X_{i+1}^2=\sum_{i=1}^n\mathbb E[X_i^2]\mathbb E[X_{i+1}^2]=n. $$ So $S_{199}$ (the quantity you are interested in) has mean $0$ and variance $199$, which means we should replace it by $\sqrt{199}$ times a standard normal random variable to get the approximation (assuming a weakly dependent central limit theorem).