Explanation

(c) Magnetic field at the centre of loop
$B = \frac{\mu_{0}}{4\pi} \bullet \frac{I.2\pi R}{R^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)$
For the wire which is looped double let radius becomes $r$
Then, $\frac{l}{2} = 2\pi r;\frac{1}{4\pi} = (r)$
$\therefore B^{''} = \frac{\mu_{0}}{4\pi}.\frac{I.2\pi \times 2}{r^{2}}$
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$\Rightarrow B^{''} = \frac{\mu_{0}}{4\pi}.\frac{I.\frac{l}{2}.2}{\left( \frac{1}{4\pi} \right)^{2}}$
$\Rightarrow B^{''} = \frac{\mu_{0}}{4\pi}.\frac{Il \times 16\pi^{2}}{l^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(iii)$
Now, $B = \frac{\mu_{0}}{4\pi}.\frac{I.l}{\left( \frac{l}{2\pi} \right)^{2}}\ \left\lbrack R = \frac{l}{2\pi} \right\rbrack$ …(iii)
Dividing equation (ii) by Equation (iii), we get
$\frac{B^{''}}{B} = \frac{\frac{\mu_{0}}{4\pi}.\frac{I.l.16\pi^{2}}{l^{2}}}{\frac{\mu_{0}}{4\pi}.\frac{Il.4\pi^{2}}{l^{2}}} \Rightarrow \frac{B^{''}}{B} = 4 \Rightarrow B^{''} = 4B$