$\because f\left(x\right)$ is continuous at $x=3$
$\therefore \underset{x\to 3}{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(x\right)=f\left(3\right)$
$2\left(3^2\right)+3=a\left(3^2\right)+b\left(3\right)+1$
$9a+3b=20$ … $\left(1\right)$
Now, $f^{\text{'}}\left(x\right)=\left\{\begin{array}{lll}4x& ;& x>3\\ 2ax+b& ;& x\le 3\end{array}\right.$
$\because f\left(x\right)$ is also differentiable at $x=3$
$\therefore 4\left(3\right)=2a\left(3\right)+b$
$6a+b=12$ … $\left(2\right)$
From $\left(1\right)$ & $\left(2\right)$, we get,

$9a=16\Rightarrow a=\frac{16}{9}, b=\frac{4}{3}$