a
(a) Case \(1\) : \({B_A} = \frac{{{\mu _0}}}{{4\pi }}.\frac{i}{r} \otimes \) 

\({B_B} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{\pi \,i}}{r}\odot\) 

\({B_C} = \frac{{{\mu _0}}}{{4\pi }}.\frac{i}{r}\odot\) 

So net magnetic field at the centre of case \(1\) 

\({B_1} = {B_B} - {B_C} - {B_A} \Rightarrow {B_1} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{\pi i}}{r}\odot\)..... \((i)\)

Case \(2\) : As we discussed before magnetic field at the centre \(O\) in this case 

\({B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{\pi i}}{r} \otimes \) ..... \((ii)\)

Case \(3\) : \({B_A} = 0\) \({B_B} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{(2\pi - \pi /2)i}}{r} \otimes \) 

\({B_C} = \frac{{{\mu _0}}}{{4\pi }}.\frac{i}{r}\odot\)

\( = \frac{{{\mu _0}}}{{4\pi }}.\frac{{3\pi i}}{{2r}} \otimes \) 

So net magnetic field at the centre of case \(3\) \({B_3} = \frac{{{\mu _0}}}{{4\pi }}.\frac{i}{r}\left( {\frac{{3\pi }}{2} - 1} \right) \otimes \)..... \((iii)\)

From equation \((i)\), \((ii)\) and \((iii)\) 

\({B_1}:{B_2}:{B_3} = \pi \odot\) : \(\pi \otimes \) \(\left( {\frac{{3\pi }}{2} - 1} \right)\, \otimes = - \frac{\pi }{2}:\frac{\pi }{2}:\left( {\frac{{3\pi }}{4} - \frac{1}{2}} \right)\)