b
Magnetic fields due to both wires will be perpendicular to each other.

\(B _1=\frac{\mu_0 i _1}{2 \pi d } \quad B _2=\frac{\mu_0 i _2}{2 \pi d }\)

\(B _{\text {net }}=\sqrt{ B _1^2+ B _2^2} \Rightarrow \frac{\mu_0}{2 \pi d } \sqrt{ i _1^2+ i _2^2}\)

\(\Rightarrow \frac{4 \pi \times 10^{-7}}{2 \pi \times(7 / \sqrt{2}) \times 10^{-2}} \times \sqrt{8^2+15^2}\left( d =\frac{7}{\sqrt{2}}\,cm \right)\)

\(\Rightarrow 68 \times 10^{-6}\,T\)