Like this?

Of course, since this is an exercise, I shouldn't disclose the code I produced the image with. ;-)
The basis is, of course, \resizebox{1}[1.6]{...}
from the graphicx
package. However, proper positioning is not really easy, but not terribly difficult either.
Well here's the commented code
\documentclass{article}
\usepackage{graphicx,calc,amsmath}
\newcommand{\ssum}{%
\mathop{%
% get the dimensions of \sum in \displaystyle
\sbox0{$\displaystyle\sum$}%
% lower the stretched \sum by its height and raise it by the height
% of the unstretched symbol; give it the height it should have
% and the right depth
\raisebox{-\height+\ht0}[\ht0][\dp0]{\scalebox{1}[1.6]{\copy0}}%
}\displaylimits
}
\newcommand{\sbrace}[2]{%
#1{% #1 is either \mathopen or \mathclose, #2 is \lbrace or \rbrace (or any delimiter)
% get the dimensions of the unstretched delimiter so that
% it covers \sum with subscript and superscript
\sbox0{$\displaystyle\left#2\vphantom{\sum_{i=1}^{n}}\right.\kern-\nulldelimiterspace$}%
% as before, place the symbol, but in this case we want its
% depth to be zero; the stretching is somewhat different
\raisebox{-\height+\ht0}[\ht0][0pt]{\scalebox{1}[1.7]{\copy0}}%
}%
}
\newcommand{\slbrace}{\sbrace{\mathopen}{\lbrace}} % stretched left brace
\newcommand{\srbrace}{\sbrace{\mathclose}{\rbrace}} % stretched right brace
\begin{document}
\[
0=\ssum_{i=1}^{n} m(X_{i},\hat{\theta}_{i})
=\ssum_{i=1}^{n} m(X_{i},\hat{\theta}_{0})
+\slbrace
\ssum_{i=1}^{n}\frac{\partial m(X_{i},\theta^{*}_{n})}{\partial \theta^{T}}
\srbrace
(\hat{\theta}_{n}-\theta_{0}).
\]
\textbf{\qquad l 2} \emph{Omdat $\hat{\theta}_{n}\xrightarrow{p}\theta^{*}_{0}$
en $\theta^{*}_{0}$ een waarde is die tussen $\hat{\theta}_{n}$ en
$\theta^{*}_{0}$ ligt, zal}
\end{document}
I'm not sure if a student is really supposed to know all that, but maybe the instructor will learn something. ;-)