Answer
a) $({\bf 2.0\times 10^{-4}}\;{\rm T})\;\hat i$
b) $({\bf 4.0\times 10^{-4}}\;{\rm T})\;\hat i$
c) $({\bf 2.0\times 10^{-4}}\;{\rm T})\;\hat i$
Work Step by Step
$$\color{blue}{\bf [a]}$$
First of all, we need to find the direction of the magnetic field of each wire at each given point.
We need to use the right-hand rule, see the figures below.
$$\sum B_a=B_1+B_2$$
We know that the magnetic field of a long straight current wire is given by
$$B=\dfrac{\mu_0I}{2\pi d}$$
$$\sum B_a=\dfrac{\mu_0I_1}{2\pi d_1}\left(\cos45^\circ\;\hat i-\sin45^\circ\hat j\right)+\dfrac{\mu_0I_2}{2\pi d_2}\left(\cos45^\circ\;\hat i+\sin45^\circ\hat j\right)$$
where $I_1=I_2=I$; and $d_1=d_2=d$,
$$\sum B_a=\dfrac{\mu_0I }{2\pi d }\left[ \left(\cos45^\circ\;\hat i-\sin45^\circ\hat j\right)+ \left(\cos45^\circ\;\hat i+\sin45^\circ\hat j\right)\right]$$
$$\sum B_a=\dfrac{\mu_0I }{2\pi d }\left[ 2 \cos45^\circ\;\hat i \right]$$
Plug the known;
$$\sum B_a=\dfrac{(4\pi\times 10^{-7})(10) }{2\pi \left(\sqrt{0.01^2+0.01^2}\right)}\left[ 2 \cos45^\circ\;\hat i \right]$$
$$\sum B_a=(\color{red}{\bf 2.0\times 10^{-4}}\;{\rm T})\;\hat i$$
$$\color{blue}{\bf [b]}$$
By the same approach,
$$\sum B_b=\dfrac{\mu_0I }{2\pi d }\left[ 2 \cos0^\circ\;\hat i \right]$$
Plug the known;
$$\sum B_b=\dfrac{(4\pi\times 10^{-7})(10) }{2\pi \left(0.01\right)}\left[ 2 \;\hat i \right]$$
$$\sum B_b=(\color{red}{\bf 4.0\times 10^{-4}}\;{\rm T})\;\hat i$$
$$\color{blue}{\bf [c]}$$
By the same approach,
$$\sum B_c=\dfrac{\mu_0I }{2\pi d }\left[ 2 \cos45^\circ\;\hat i \right]$$
Plug the known;
$$\sum B_c=\dfrac{(4\pi\times 10^{-7})(10) }{2\pi \left(\sqrt{0.01^2+0.01^2}\right)}\left[ 2 \cos45^\circ\;\hat i \right]$$
$$\sum B_c=(\color{red}{\bf 2.0\times 10^{-4}}\;{\rm T})\;\hat i$$
