You're trying to find constants [tex]a_0,a_1,a_2[/tex] such that [tex]\hat y=a_0+a_1\hat x+a_2{\hat x}^2[/tex]. Equivalently, you're looking for the least-square solution to the following matrix equation.
[tex]\underbrace{\begin{bmatrix}1&6&6^2\\1&3&3^2\\\vdots&\vdots&\vdots\\1&9&9^2\end{bmatrix}}_{\mathbf A}\underbrace{\begin{bmatrix}a_0\\a_1\\a_2\end{bmatrix}}_{\mathbf x}=\underbrace{\begin{bmatrix}100\\110\\\vdots\\70\end{bmatrix}}_{\mathbf b}[/tex]
To solve [tex]\mathbf{Ax}=\mathbf b[/tex], multiply both sides by the transpose of [tex]\mathbf A[/tex], which introduces an invertible square matrix on the LHS.
[tex]\mathbf{Ax}=\mathbf b\implies\mathbf A^\top\mathbf{Ax}=\mathbf A^\top\mathbf b\implies\mathbf x=(\mathbf A^\top\mathbf A)^{-1}\mathbf A^\top\mathbf b[/tex]
Computing this, you'd find that
[tex]\mathbf x=\begin{bmatrix}a_0\\a_1\\a_2\end{bmatrix}\approx\begin{bmatrix}121.119\\-3.786\\-0.175\end{bmatrix}[/tex]
which means the first choice is correct.
