The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is _______.

According to Einstein’s photoelectric equation,
$K_{\max }=hv-{\varphi }_0$
Where the symbols have their usual meaning.
But  $K_{\max }=e{V}_s$ where $V_s$ is the stopping potential
    $V_s=\frac{h}{e}v-\frac{{\varphi }_0}{e}$
Thus, the graph between $V_s-v$ is a straight line.
Compare the above relative with  $y=mx+C$
$\therefore$  Slope of $V_s-v$ graph $=\frac{h}{e}$
It is same for both the metals.

 $\therefore$   Ratio of the slopes = 1