Answer
${\bf 7.0}\;\rm A$
Work Step by Step
According to Ampere's law, the line integral of the magnetic field around a closed path is given by
$$\oint \vec B\cdot d\vec s=\mu_0I_{through}$$
where $ I_{through}=I_1+I_3-I_2$ is the current through the area bounded by the integration path.
$$\oint \vec B\cdot d\vec s=\mu_0(I_1+I_3-I_2)$$
We can see, from the given graph, and from the right-hand rule, that the magnetic field of $I_2$ is the only field that opposes the path direction.
Solving for $I_3$;
$$I_3=\dfrac{\oint \vec B\cdot d\vec s}{\mu_0}-I_1+I_2$$
Plug the known;
$$I_3=\dfrac{(3.77\times 10^{-6})}{(4\pi \times 10^{-7})}-2+6$$
$$I_3=\color{red}{\bf 7.0}\;\rm A$$
