Solution :
given $ n=16$
$r= 10\;cm=0.1 \;m$
$I= 0.75 \;A$
$B= 5 \times 10^{-2} T$
$ f= \large\frac{2}{s}$
$M= nIA$
$\qquad= 16 \times 0.75 \times \pi (0.1)^2$
$\qquad= 0.377\;J/T$
Frequency $f= \large\frac{1}{2\pi} \sqrt {\large\frac{M \times B}{I}}$
(ie) $f^2 =\large\frac{1}{4 \pi ^2} \frac{MB}{I}$
$\therefore I= \large\frac{1}{4 \pi ^2} \frac{MB}{f^2}$
$\qquad = \large\frac{ 0.377 \times 5 \times 10^{-2}}{4 \times (3.14)^2 \times 2 \times 2 }$
$\qquad= 1.2 \times 10^{-4}\;kg\; m^2$
Hence the normal of inertia of the coil $ 1.2 \times 10^{-4}\;kg\; m^2$
Answer : $ 1.2 \times 10^{-4}\;kg\; m^2$