Three identical rods each of mass ‘M’ and length ‘L’ are joined to form a symbol ‘H’. The moment of inertia of the system about one of the sides of ‘H’ is

The given situation can be shown as

Let us take the moment of inertia of the system about rod $R_1$ then total moment of inertia is
For rod $R_1{l}_1=0$
For rod $R_2$ , using perpendicular axis theorem,
$l_2=\frac{M{L}^2}{3}$
For rod $R_3$ , using parallel axis theorem,
$l_3=l_{CM}+l_{\left(atL\right)}=0+M{L}^2=M{L}^2$
Now, putting the values of $l_1,l_2$ and $l_3$ in Eq. (i), we get
$l_{\tau }=0+\frac{M{L}^2}{3}+M{L}^2\Rightarrow l_{\tau }=\frac{4M{L}^2}{3}$