Explanation

(a) Magnetic field at the centre of current carrying coil is given by
$B = \frac{\mu_{0}}{4\pi}\frac{2\pi Ni}{r}$
$\Rightarrow B \propto \frac{N}{r}$
$\frac{B_{1}}{B_{2}} = \frac{N_{1}}{N_{2}} \times \frac{r_{2}}{r_{1}}$
In second condition, a loop having two turns is prepared from the same wire.
So, $N_{1} = N$
$N_{2} = 2N$
$r_{1} = r$
$r_{2} = 2r$
$\Rightarrow \frac{B}{B_{2}} = \frac{1}{2} \times \frac{r/2}{r} = \frac{1}{4} \Rightarrow B_{2} = 4B$