Answer:
By using theorem of parallel axes, \[{{I}_{end}}={{I}_{CM}}+M{{\left( \frac{l}{2} \right)}^{2}}=\frac{1}{12}M{{l}^{2}}+\frac{1}{4}M{{l}^{2}}=\frac{1}{3}M{{l}^{2}}\]
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