Answer
${\bf 7.85\times 10^{-5}}\;\rm T\tag{Into the page}$
Work Step by Step
We have here 4 wires, the two straight wires plus the two semi-circular wires.
The magnitudes of the magnetic fields due to the two straight wires, at point P, are zeros since the angles are zeros.
Now it is obvious that the net magnetic field at point P is due to the semi-circular wires only.
So, the net magnetic field at point P is given by
$$B_{net}=B_{\rm big}-B_{\rm small}$$
where big is for the bigger circle while small is for the smaller one.
We can call the big one as 1 and the small one as 2.
Thus,
$$B_{net}=B_{ 1}-B_{ 2}\tag 1$$
where, according to the right-hand rule, the direction of the magnetic field of 1 is out of the page [which we considered as the positive direction] while the direction of the magnetic field of 2 is into the page. That's why we put a negative sign in front of $B_2$.
Recalling that the magnitude of the magnetic field of a very short segment $ds$ is given by
$$dB=\dfrac{\mu_0 I ds}{4\pi R^2} $$
where for a small segment of the arc $ds=Rd\theta$ where $\theta$ in radians.
So,
$$dB=\dfrac{\mu_0 I Rd\theta}{4\pi R^2} =\dfrac{\mu_0 I d\theta}{4\pi R } $$
Taking the integral,
$$\int_0^BdB=\int_0^\theta\dfrac{\mu_0 I d\theta}{4\pi R } $$
$$ B=\dfrac{\mu_0 I }{4\pi R } \int_0^{\pi} d\theta $$
$$ B=\dfrac{\mu_0 I \pi }{4\pi R } $$
$$B=\dfrac{\mu_0 I }{4 R }\tag 2 $$
Plug into (1),
$$B_{net}=\dfrac{\mu_0 I_1 }{4 R_1 }-\dfrac{\mu_0 I_2 }{4 R_2 }$$
$$B_{net}=\dfrac{\mu_0 }{4 } \left[ \dfrac{ I_1 }{ R_1 }-\dfrac{I_2 }{ R_2 }\right]$$
Plug the known;
$$B_{net}=\dfrac{(4\pi\times 10^{-7})}{4 } \left[ \dfrac{ 5}{ 0.02}-\dfrac{5}{ 0.01}\right]$$
$$B_{net}=-\color{black}{\bf 7.85\times 10^{-5}}\;\rm T$$
The negative sign indicates the field's direction.
$$B_{net}= \color{red}{\bf 7.85\times 10^{-5}}\;\rm T\tag{Into the page}$$
