Answer
a) $ B_{\rm net}= { \mu_0 NI }/{ (\frac{5}{4} )^{3/2}R } $
b) ${\bf 1.8\times 10^{-4}}\;\rm T$
Work Step by Step
$$\color{blue}{\bf [a]}$$
Since the two coils are identical, parallel to each other on the same axis, they each have the same $N$ turns, and are carrying an equal current $I$ in the same direction, the net magnetic field between them is the sum of both fields.
$$B_{\rm net}=B_1+B_2=2B_{\rm coil}\tag 1$$
We know that the magnitude of the magnetic field at some point on the axis of a current loop is given by
$$B=\dfrac{\mu_0 I R^2}{2(z^2+R^2)^{3/2}} $$
and here $z=R/2$ since we need to find the net field between the two coils and the distance between them is $R$, and each coil contains $N$ turns. So
$$B_{\rm coil}=\dfrac{\mu_0 NI R^2}{2(\frac{R^2}{4}+R^2)^{3/2}}=\dfrac{\mu_0 NI R^2}{2(\frac{1}{4}+1)^{3/2}R^3 }$$
$$B_{\rm coil}= \dfrac{\mu_0 NI }{ 2(\frac{1}{4}+1)^{3/2}R }$$
Plug into (1);
$$B_{\rm net}= \dfrac{2\mu_0 NI }{ 2(\frac{1}{4}+1)^{3/2}R } $$
$$\boxed{B_{\rm net}= \dfrac{ \mu_0 NI }{ (\frac{5}{4} )^{3/2}R } }$$
$$\color{blue}{\bf [b]}$$
Just plug the given into the boxed formula above,
$$ B_{\rm net}= \dfrac{ (4\pi\times 10^{-7})(10)(1) }{ (\frac{5}{4} )^{3/2}(0.05) } $$
$$ B_{\rm net}=\color{red}{\bf 1.8\times 10^{-4}}\;\rm T$$
