Answer
$a=1$, $b=1$ and $c=0$.
Work Step by Step
$$y=f(x)=ax^2+bx+c$$
The derivative of $f(x)$ is $$f'(x)=2ax+b$$
Since the curve $f(x)$ is tangent to $y=x$ at the origin, $f'(0)$ is equal to the slope of $y=x$, which is $1$: $$f'(0)=1$$
$$2a\times0+b=1$$
$$b=1$$
Also, $f(x)$ being tangent to $y=x$ at the origin means that the graph of $f(x)$ passes through the point $(0,0)$: $$a\times0^2+b\times0+c=0$$
$$c=0$$
Finally, since $f(x)$ passes through $(1,2)$, we have: $$a\times1^2+b\times1+c=2$$
$$a+b+c=2$$
Substitute $b=1$ and $c=0$ here: $$a+1+0=2$$
$$a=1$$
