
Since wires are connected in series, therefore some current will flow in the two wires.
we know that,
\(I = neAv_d \)
\(\therefore \ I_A = I_B \Rightarrow n_AeA_A \times \left(\vec v_d\right)_A = n_BeA_B \times \left(\vec v_d\right)_B\)
Given \(n_A = 1.5 n_B, \ A_A = A_B\)
\(\therefore \ 1.5n_B \times \left(\vec v_d\right)_A = n_B \times \left(\vec v_d\right)_B\)
\(\frac{\left(\vec v_d\right)_A}{\left(\vec v_d\right)_B} = \frac{1}{1.5} = \frac{2}{3}\)