The M.I. of a thin rod about a normal axis through its centre is I. It is bent at the centre such that, the two parts are perpendicular to each other and perpendicular to the axis. The M.I. of the system about the same axis will be

According to the parallel axis theorem, the M.I. of any body about any axis is equal to the product of the mass of the body and the square of the distance between the two axes plus the M.I. of the body about a parallel axis that passes through the centre of gravity.

From the question,

$\mathrm{I}=\frac{{\mathrm{ml}}^2}{12}$

For the bent shape

$\mathrm{I}\text{'}=\frac{\mathrm{m}}{2}\frac{{\left({\displaystyle \frac{\mathrm{l}}{2}}\right)}^2}{3}+\frac{\mathrm{m}}{2}\frac{{\left({\displaystyle \frac{\mathrm{l}}{2}}\right)}^2}{3}=\mathrm{I}$

Hence the correct answer is I.