calculate the internally studentized residual 
it says that ...an ordinary residual divided by an estimate of its standard deviation $s(e_{i})$
As we can see from the example that mean for four residuals is 0, so $s(e_{i})=\sqrt{\frac{(-0.2-0)^2+(0.6-0)^2+(-0.6)^2+(0.2-0)^2}{4-1}}=\sqrt{\frac{0.8}{3}}\neq\sqrt{0.4(1-0.7)}$
where did I get it wrong?
 A: You are only diving by $4-1$, when presumably you should be dividing by $N-p$, where $N$ is the number of observations and $p$ is the number of parameters you've estimated.  I'm assuming this is a simple linear regression model, in which case you've estimated two parameters:  $\beta_0$ and $\beta_1$, so $p=2$.  Then, 
$MSE=\sqrt{\frac{(-0.2-0)^2+(0.6-0)^2+(-0.6)^2+(0.2-0)^2}{4-2}}=\sqrt{\frac{0.8}{2}}=\sqrt{0.4}$
Now, $s(e_i)=\sqrt{MSE(1-h_{ii})}$, and for $i=1$, $h_{11}=0.7$ (given), so 
$s(e_1)=\sqrt{0.4(1-0.7)}$.  Finally,
$r_1=$$-0.2\over{{\sqrt{0.4(1-0.7)}}}$
Additional Comments to Your Follow-up Question
You are calculating the standard error of the residuals from a regression, so the formula is a little more complicated than that of the usual standard deviation of a number.  To see how the standard error is obtained, note:
$e=Y-\hat{Y}=Y-X\hat{\beta}=Y-X(X^{\prime}X)^{-1}X^{\prime}Y=Y-HY=Y(I-H)$.  Now, $Var(e)=Var[Y(I-H)]=Var(Y)(I-H)^2=Var(Y)(I-H)$, because $(I-H)$ is idempotent [i.e. $(I-H)(I-H)=(I-H)$].  Now, $Var(Y)=\sigma^2I$.  So, $Var(e)=\sigma^2(I-H)$.  To pick out the $i$th residual, $e_i$, we need the ($i,i$) diagonal element of $(I-H)$ ($1-h_{ii}$), so $Var(e_i)=\sigma^2(1-h_{ii})$.  the variance for the first residual is $Var(e_1)=\sigma^2(1-h_{11})=\sigma^2(1-h_{11})$ and so the corresponding standard error is $se(e_1)=\sqrt{\sigma^2(1-h_{11})}$  Since we don't know $\sigma^2$, we estimate it from the data and that is estimated by $MSE$ as given in the original answer.
