Answer: $\Theta(n^3)$
The Loop Equation can be written as:-
$\sum_{i=1}^{n}\sum_{j=1}^{i}\sum_{k=1}^{j} c$
$=\sum_{i=1}^{n}\sum_{j=1}^{i}(j)$
$=\sum_{i=1}^{n}(i(i+1)/2)$
$=\sum_{i=1}^{n}(i^2+i)/2)$
$=(1/2)\sum_{i=1}^{n}(i^2+i)$
$=(1/2)\{[(n(n+1)(2n+1)/6] + [(n(n+1)/2)]\}$
$=\Theta(n^3)$