Variance of $\hat{\beta _0}$ in case of pure homoskedasticity Stock and Watson express the variance of $\hat{\beta _0}$ like 
$\hat{\sigma }^2_\hat{\beta _0}=\frac{E({X_{i}}^{2})}{n\sigma _{X}^{2}}\sigma ^{2}$, but starting from variance of $\hat{\beta _1}=\frac{\sigma^{2}}{n\sigma_{X}^{2}}$ i proved only that $\hat{\sigma }^2_\hat{\beta _0}=\frac{1}{n}\sigma^{2}(1+\frac{\bar{X}^2}{\sigma_{X}^{2}})$, that is the same that is showed here. 
How can i prove that are similar forms? 
 A: Let us depart from the heteroskedastic result 
$$
\sigma^2_{\hat\beta_0}=\frac{1}{n}\frac{var(H_iu_i)}{[E(H_i^2)]^2}$$
which Stock & Watson specialize to homoskedasticity here. Here, 
$$
H_i=1-\left(\frac{\mu_X}{E(X_i^2)}\right)X_i.
$$
We may always write
$$
var(H_iu_i)=E(H_i^2u_i^2)-[E(H_iu_i)]^2
$$
Under the maintained assumption of correct specification, $E(u_i|X_i)=0$ and hence, by the law of iterated expecations, as $H_i$ only depends on $X_i$,
$$
E(H_iu_i)=E(H_iE(u_i|X_i))=0
$$
Hence,
$$
var(H_iu_i)=E(H_i^2u_i^2)
$$
Again by the LIE,
$$
E(H_i^2u_i^2)=E(H_i^2E(u_i^2|X_i))
$$
which reduces to
$$
E(H_i^2E(u_i^2|X_i))=\sigma^2_uE(H_i^2) 
$$
under homoskedasticity. So
$$
\sigma^2_{\hat\beta_0}=\frac{1}{n}\frac{\sigma^2_uE(H_i^2)}{[E(H_i^2)]^2}=\frac{1}{n}\frac{\sigma^2_u}{E(H_i^2)}
$$
Let us study 
\begin{eqnarray*}
E(H_i^2)
&=&1-2\frac{\mu_XE(X_i)}{E(X_i^2)}+\frac{\mu_X^2}{(E(X_i^2))^2}E(X_i^2)\\
&=&1-2\frac{\mu_X^2}{E(X_i^2)}+\frac{\mu_X^2}{(E(X_i^2))^2}E(X_i^2)\\
&=&1-2\frac{\mu_X^2}{E(X_i^2)}+\frac{\mu_X^2}{E(X_i^2)}\\
&=&\frac{E(X_i^2)-2\mu_X^2+\mu_X^2}{E(X_i^2)}\\
&=&\frac{E(X_i^2)-\mu_X^2}{E(X_i^2)}\\
&=&\frac{\sigma^2_X}{E(X_i^2)},
\end{eqnarray*}
which gives the desired result.
