Answer
$2.4 \times 10^{-2}N/m$
Work Step by Step
As explained in the text, in figure 10–34 there is both a top and a bottom surface.
$$\gamma = \frac{F}{2 \mathcal{l}}$$
$$\gamma=\frac{3.4 \times 10^{-3}N}{2(0.070m)}=2.4 \times 10^{-2}N/m$$
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