Answer
The percentage error is $~~0.2275\%$
Work Step by Step
In part (a), we found that $I = 0.008352~kg~m^2$
We can find an approximation for $I$:
$I' \approx \frac{1}{12}ML^2$
$I' \approx \frac{1}{12}(0.1000~kg)(1.0000~m)^2$
$I' \approx 0.008333~kg~m^2$
We can find $\frac{I'}{I}$:
$\frac{I'}{I} = \frac{0.008333~kg~m^2}{0.008352~kg~m^2} = 0.997725$
Note that $~~1.0000-0.997725 = 0.002275$
The percentage error is $~~0.2275\%$
