Answer
$$V_A>V_B>V_C$$
Work Step by Step
From the equation,
$$eV = h\nu - W_0$$
where V is the stopping potential $\nu $ is the frequency of incident light and $W_0$ is the work function of the metal.
From this equation, we can conclude that for same $\nu$ larger the $W_0$, lesser the stopping potential $V$
Therefore,
$$V_A>V_B>V_C$$
