Order-of-growth of selection sort is quadratic, it takes 2secs to sort 100 elements, what will be the expected completion time to sort 200 elements?
First off, that's a less than wonderful question.

Second: If the running time is indeed quadratic (with no linear term), the best answer is 8 seconds, not 4 seconds.  Suppose the running time is $T(n) = c \cdot n^2$ for some constant $c$.  Then we know $T(100) = $ 2 seconds.  One can show that

$$T(200) = c \cdot 200^2 = c \cdot 4 \cdot 100^2 = 4 \times T(100),$$

so it follows that $T(200) =$ 8 seconds.

That said, I still don't like the question very much.  Usually when we say that an algorithm has quadratic running time, we mean that its running time is in $\Theta(n^2)$.  That means the running time could be something like $T(n) = c_1 \cdot n^2 + c_2 \cdot n$.  If the running time looks like that, then we don't have enough information to conclude what the running time will be for $n=200$.