A straight piece of wire 5 cm long is placed horizontally on the surface of water and gently pulled up. It is found that a force of 7×10^–3 N in addition to weight of wire is required for that purpose, the surface tension of water is 

The water molecules will draw the wire from both sides along its length.
5 cm is the wire's length. 

$\mathrm{L}=2\times \mathrm{l}=2\times 5 \mathrm{cm}=10 \mathrm{cm}$ is the length of the wire on which surface tension is acting.

The extra 728 dyne force is necessary to balance the force that surface tension has on the wire.

$\mathrm{F}=\mathrm{T}\times \left(2l\right) \Rightarrow \mathrm{T}(2\times 5\times {10}^{-2})=7\times {10}^{-3}$

$\Rightarrow \mathrm{T}=\frac{7\times {10}^{-2}\times {10}^{-1}}{10\times {10}^{-2}}=7\times {10}^{-2}\mathrm{N}/\mathrm{m}.$

Hence the correct answer is $7\times {10}^{-2}\mathrm{N}/\mathrm{m}.$