The following taylor expansion of the function $f(x)$, requires $f(x)$ to have a derivative up to what order? $$ f(x)=f(0)+f'(0)x+f''(0)x^2/2+\mathcal{O}(x^3)$$
My solution: Based on the Taylor's Theorem
$f(x)=f(a)+f'(a)(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+\cdots +{\frac {f^{(k)}(a)}{k!}}(x-a)^{k} + \ R_{k}(x)$
where $R_{k}(x)={\frac {f^{(k+1)}(\xi _{L})}{(k+1)!}}(x-a)^{k+1}$ and $ξ_L$ is between a and x. Therefore, the expansion in the question requires $f(x)$ to have up to the third derivative.
Is my solution correct? Is there a way that we need only up to the 2nd derivative?
tl;dr: as you expected, existence of the third derivative at $0$ is necessary and sufficient.
By your argument, differentiable thrice at $0$ is sufficient. Now, let us show it is necessary.
For that, let $f\colon\mathbb{R}\to\mathbb{R}$ be the even function defined by $f(x) = \lvert x\rvert^{5/2}$. It is a simple matter to check that it is twice differentiable; so that it $f\in\mathcal{D}^2$ were a sufficient condition, then we would have $|x|^{5/2} = O(x^3)$ around $0$ — which is clearly false.
As a side note: as mentioned in the comments, if the question were to find a condition on $f$ ensuring that $$ f(x) = f(0)+f'(0)x+\frac{f''(0)}{2}x^2 + o(x^2) $$ instead, then indeed $f$ being twice differentiable at $0$ would be enough. But the remainder being $O(x^3)$ is a strictly stronger condition than $o(x^2)$.