Answer
$$
d=\frac{2}{3} I
$$
Work Step by Step
$$
\begin{array}{ll}
(\leftrightarrows) \quad & m\left(v_{G i x}\right)_1+\Sigma \int F_x d t=m\left(v_{G x}\right)_2 \\
& 0+P(t)=m\left(v_G\right)_x \\
(+\uparrow) \quad & m\left(v_G\right)_1+\Sigma \int F_y d t=m\left(v_G\right)_2 \\
& 0+0=m\left(v_G\right)_y \\
& \left(H_G\right)_1+\Sigma / M_G d t=\left(H_G\right)_2 \\
& 0+P(t)\left(\frac{L}{2}\right)=\frac{1}{12} m L^2 \omega \\
& v_G=y \omega \\
& m\left(v_G\right)_x\left(\frac{L}{2}\right)=\frac{1}{12} m L^2 \omega \\
& \left(v_G\right)_x=\frac{L}{6} \omega \\
& y=\frac{L}{6} \\
& d=\frac{L}{2}+\frac{L}{6}=\frac{2}{3} L
\end{array}
$$
