A $4\hspace{0.17em}A$ current carrying loop consists of three identical quarter circles of radius $5 cm$ lying in the positive quadrants of the $x-y, y-z and z-x$ planes with their centres at the origin joined together, value of $\overrightarrow{B}$ at the origin is 

$$\begin{gathered} As \overrightarrow{B}=\overrightarrow{B_{xy}}+{\overrightarrow{B}}_{yz}+{\overrightarrow{B}}_{zx} .....\left(i\right) \\ where {\overrightarrow{B}}_{xy}=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{I}{R}\theta \hat{k}, {\overrightarrow{B}}_{yz}=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{I}{R}\theta \hat{i}, {\overrightarrow{B}}_{zx}=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{I}{R}\theta \hat{j} \end{gathered}$$

Substituting these values in (i) we get, 

$$\begin{gathered} \overrightarrow{B}=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{I}{R}\theta \left(\hat{i}+\hat{j}+\hat{k}\right) \\ Here, \theta =\frac{\mathrm{\pi }}{2}, I=4 A and R=5 cm=5\times {10}^{-2}m \\ \therefore \overrightarrow{B}=\frac{{\mu }_0}{4\mathrm{\pi }}\times \frac{4}{5\times {10}^{-2}}\times \frac{\mathrm{\pi }}{2}\left(\hat{i}+\hat{j}+\hat{k}\right)=10{\mu }_0\left(\hat{i}+\hat{j}+\hat{k}\right)T \end{gathered}$$