$A$ be the area of cross section,

$e$ is the charge on electron and

$I$ is the current flowing through both the wires

Hence, $I=n_{1} e A V_{d_{1}}=n_{1} e A V_{d_{2}}$

$\therefore n_{1} V_{d_{1}}=n_{2} V_{d_{2}}$

$\therefore \frac{V_{d 1}}{V_{d_{2}}}=\frac{n_{2}}{n_{1}}$

$\therefore \frac{V_{d 1}}{V_{d_{2}}}=\frac{4}{1}$