\(Y = \frac{{FL}}{{A\Delta L}} = \frac{{4FL}}{{\pi {D^2}\Delta L}}\,\,or\,\,\Delta L = \frac{{4FL}}{{\pi {D^2}Y}}\)

Where \(F\) is the force applied, \(L\) is the length, \(D\) is the diameter and \(\Delta L\) is the extension of the wire respectively. 

As each wire is made up of same material therefore their \(Young's\) modulus is same for each wire.

Foe all the four wires, \(Y,F\,(=tension)\) are the same.

\(\therefore \Delta L \propto \frac{L}{{{D^2}}}\)

\(In\,\,\left( a \right)\,\,\frac{L}{{{D^2}}} = \frac{{200\,cm}}{{{{\left( {0.2\,cm} \right)}^2}}} = 5 \times {10^3}\,c{m^{ - 1}}\)

\(In\,\,\left( b \right)\,\,\,\frac{L}{{{D^2}}} = \frac{{300\,cm}}{{{{\left( {0.3\,cm} \right)}^2}}} = 3.3 \times {10^3}\,c{m^{ - 1}}\)

\(In\,\,\left( c \right)\,\,\frac{L}{{{D^2}}} = \frac{{50\,cm}}{{{{\left( {0.05\,cm} \right)}^2}}} = 20 \times {10^3}\,c{m^{ - 1}}\)

\(In\,\,\left( d \right)\,\,\frac{L}{{{D^2}}} = \frac{{100\,cm}}{{{{\left( {0.1\,cm} \right)}^2}}} = 10 \times {10^3}\,\,c{m^{ - 1}}\)