Answer
${\bf 4.14\times 10^{-4}}\;\rm T\tag{Into the page}$
Work Step by Step
We have here 3 wires, the two straight wires plus the loop.
We can assume that the two horizontal straight wires as one very long wire since the direction of the current in both of them is the same and since they are overlapped.
According to the right-hand rule, the direction of the magnetic field due to this wire is into the page which we chose to be our negative $z$-direction, and the direction of the magnetic field due to the loop wire is also into the page
Thus,
$$B_{net}=-(B_{ \rm loop\;center}+B_{ \rm wire})\tag 1 $$
We know that the magnitude of the magnetic field at the center of the axis of a current loop is given by
$$B=\dfrac{\mu_0 I R^2}{2(z^2+R^2)^{3/2}} $$
At the center of the loop $z=0$, so
$$B_0=\dfrac{\mu_0 I R^2}{2R^3} $$
$$B_0=\dfrac{\mu_0 I_1}{2R}\tag 2$$
and we know that the magnitude of the magnetic field due to a wire is given by $$B=\dfrac{\mu_0I}{2\pi d}\tag 3$$
Plug (2) and (3) into (1);
$$B_{net}=-\left[ \dfrac{\mu_0 I_1}{2R}+\dfrac{\mu_0I_2}{2\pi d}\right]$$
where $d$ here is equal to the radius of the loop, $d=R$.
And $I_1=I_2$
$$B_{net}=-\left[ \dfrac{\mu_0 I }{2R}+\dfrac{\mu_0I}{2\pi R}\right]$$
$$B_{net}= \dfrac{\mu_0 I }{2R}\left[-1-\dfrac{1} {\pi }\right]$$
Plug the known;
$$B_{net}=\dfrac{(4\pi\times 10^{-7})(5) }{2(0.01)}\left[-1-\dfrac{1} {\pi }\right]$$
$$B_{net}=-\color{black}{\bf 4.14\times 10^{-4}}\;\rm T$$
The negative sign indicates the field's direction.
$$B_{net}= \color{red}{\bf 4.14\times 10^{-4}}\;\rm T\tag{Into the page}$$
