c
It should be mentioned, \(10 \; cm\) wire is part of long wire.

Force experienced by unit length of wire

\(=\frac{\mu_{0} I _{1} I _{2}}{2 \pi d }, I _{1}= I _{2}=5 \; A\)

Force experienced by wires of length \(10 \; cm\)

\(=\frac{\mu_{0} I _{1} I _{2}}{2 \pi d } \times 10 \times 10^{-2}\)

\(10^{-5}=\frac{2 \times 10^{-7} \times 5 \times 5}{ d } \times 10 \times 10^{-2}\)

\(d =50 \times 10^{-3} \; m\)

\(d =50 \times 10^{-1} cm =5 \; cm\)