Answer
${\bf 23.0}\;\rm A\tag{Into the page}$
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.
$$\oint \vec B\cdot d\vec s=\mu_0( I_3-I_2)$$
We can see, from the given graph, and from the right-hand rule, that the magnetic field of $I_2$ opposes the path direction inside the closed path.
We don't know the direction of the current $I_3$, but if the final result was positive, then it must be into the page, and vice versa.
Solving for $I_3$;
$$I_3=\dfrac{\oint \vec B\cdot d\vec s}{\mu_0} +I_2$$
Plug the known;
$$I_3=\dfrac{(1.38\times 10^{-5})}{(4\pi \times 10^{-7})} +12$$
$$I_3=\color{red}{\bf 23.0}\;\rm A\tag{Into the page}$$
