Explanation

(c) Magnetic field at the centre of a current carrying loop is given by
$\ \ \ \ \ \ \ \ \ \ \ B = \ \frac{\mu_{0}ni}{2r}$
Here, $n =$ no. of turns in loop
$\ \ \ \ \ \ \ \ \ \ \ \ i = current,\ r_{1} = radius\ of\ loop,\ \ \ r_{1} = r$
For $\ \ n = 1\ turn$
$\ \ \ \ \ \ \ \ \ \ B = \frac{\mu_{0}i}{2r_{1}}$ …(i)
When $n = 2$ turns and radius $r_{2} = \frac{r}{2},\ i_{2} = i$
$\ \ \ \ \ \ \ \ \ \ B_{2} = \frac{\mu_{0} \times 2 \times i}{2\left( \frac{r}{2} \right)}$
or $\ \ \ B_{2} = \frac{2\mu_{0}i\ \times \ 2}{2r}$ …(ii)
Now, from Eqs. (i) and (ii)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{B_{2}}{B} = 4$
Hence, $B_{2} = 4B$