Answer
$\bf {Sodium}$
Work Step by Step
We know that the stopping potential is given by
$$V_{\rm stop}=\dfrac{\dfrac{hc}{\lambda}-E_0}{e} \tag 1$$
To find the metal from which the cathode is made, we need to find $E_0$ for this metal.
We have here 2 cases, $\lambda_1=400$ nm, and $\lambda)_2=300$ nm.
where we know that
$$(V_{\rm stop})_1=0.257(V_{\rm stop})_2$$
Using (1);
$$ \dfrac{\dfrac{hc}{\lambda_1}-E_0}{ \color{red}{\bf\not} e} =0.257\left[\dfrac{\dfrac{hc}{\lambda_2}-E_0}{ \color{red}{\bf\not} e}\right]$$
$$\dfrac{hc}{\lambda_1}-E_0=\dfrac{0.257hc}{\lambda_2}-0.257E_0$$
Hence,
$$0.743E_0=\dfrac{hc}{\lambda_1}-\dfrac{0.257hc}{\lambda_2}$$
$$0.743E_0=hc\left[\dfrac{1}{\lambda_1}-\dfrac{0.257 }{\lambda_2}\right]$$
$$E_0=\dfrac{hc}{0.743 }\left[\dfrac{1}{\lambda_1}-\dfrac{0.257 }{\lambda_2}\right]$$
Plug the known;
$$E_0=\dfrac{(6.626\times 10^{-34})(3\times 10^8)}{0.743 }\left[\dfrac{1}{(400\times 10^{-9})}-\dfrac{0.257 }{(300\times 10^{-9})}\right]$$
$$E_0=\bf 4.39652\times 10^{-19}\;\rm J=\color{red}{\bf 2.75}\;\rm eV$$
Therefore, from Table 38.1, it is the $\bf \underline{Sodium}$.
