(d) Magnetic field induction at the centre of circular coil carrying current is $\ B = \frac{\mu_{0}}{4\pi}\frac{2\pi ni}{r}\ ie,\ B\ \propto n/r$
But, $2\pi r = 3 \times 2\pi r_{1}\ \ \ \ \ \ \ \ or\ \ r_{1} = r/3$
So, $\frac{B_{1}}{B} = \frac{n_{1}}{r_{1}} \times \frac{r}{n} = \frac{3 \times r}{(r/3) \times 1} = 9$