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 $10 \; cm$
$=\frac{\mu_{0} I _{1} I _{2}}{2 \pi d } \times 10 \times 10^{-2}$
$10^{-5}=\frac{2 \times 10^{-7} \times 5 \times 5}{ d } \times 10 \times 10^{-2}$
$d =50 \times 10^{-3} \; m$
$d =50 \times 10^{-1} cm =5 \; cm$
