Answer
$(a)\space 14.3\space gm^{2}$
$(b)\space 0.064\%$
Work Step by Step
(a) Here we use the equation $I=\frac{1}{12}M(a^{2}+b^{2})$ for a flat plate. Where $I $ - Rotational inertia, a - width of the rod, b - Length of the rod.
Let's plug known values into this equation.
$I=\frac{1}{12}\times172\space g(2.54^{2}\times10^{-4}m^{2}+1\space m^{2})$
$I=14.3\space gm^{2}$
(b) In part (a) we neglect dimension 2.54 cm from our calculation due to its small value with respect to width. So,
Percentage of error $=\frac{\frac{172\space g}{12}\times2.54^{2}\times10^{-4}\space m^{2}}{14.3\space gm^{2}}\times100\%$
$Error \space \%=0.064\%$
