a
Let \('L'\) and \(A\) be lenght and area of cross section of each wire. In order to have the lower ends of the wires to be at the same level \((i.e.,\,same\,elongation\,is\,produced\,in\,both\,wiers)\) let weights \(W_s\) and \(W_b\) are added to steel and brass wires respectively. Then, By definition of \(Young's\) modulus, the elongation produced in the steel wire is 

 \(\Delta {L_s} = \frac{{{W_s}L}}{{{Y_s}A}}\)                          \(\left( {asY = \frac{{W/A}}{{\Delta L/L}}} \right)\)

and that in the brass wire is   \(\Delta {L_b} = \frac{{{W_b}L}}{{{Y_b}A}}\)

But \(\Delta {L_s} = \Delta {L_b}\)                           \((given)\)

\(\therefore \frac{{{W_s}L}}{{{Y_s}A}} = \frac{{{W_b}L}}{{{Y_b}A}}\,\,or\,\,\frac{{{W_s}}}{{{W_b}}} = \frac{{{Y_s}}}{{{Y_b}}}\)

\(As\,\frac{{{Y_s}}}{{{Y_b}}} = 2\)                            \((given)\)

\(\therefore \frac{{{W_s}}}{{{W_b}}} = \frac{2}{1}\)