Answer
$\color{blue}{\frac{1}{18}}$
Work Step by Step
$\begin{align*}
E(W) &= E\left(\left(Y-\frac{2}{3}\right)^2\right) \\
&= E\left( Y^2 - \frac{4}{3}Y + \frac{4}{9} \right) \\
&= E\left(Y^2\right) - \frac{4}{3}E(Y) + \frac{4}{9} \\
&= \int_{\Omega_Y} y^2\cdot f_Y(y)\ dy -\frac{4}{3}\int_{\Omega_Y} y\cdot f_Y(y) + \frac{4}{9} \\
&= \int_0^1 y^2\cdot (2y)\ dy -\frac{4}{3}\int_0^1 y\cdot (2y)\ dy + \frac{4}{9} \\
&= 2\int_0^1 y^3\ dy - \frac{8}{3}\int_0^1 y^2\ dy + \frac{4}{9} \\
&= 2\left( \frac{y^4}{4}\ \right\vert_0^1 - \frac{8}{3}\left(\frac{y^3}{3}\ \right\vert_0^1 + \frac{4}{9} \\
&= \frac{1}{2}\left(1^4-0^4\right) - \frac{8}{9}\left(1^3-0^3\right) + \frac{4}{9} \\
&= \frac{1}{2} -\frac{8}{9} + \frac{4}{9} \\
&= \frac{9 - 16 + 8}{18} \\
\color{blue}{E(W)}\ &\color{blue}{= \frac{1}{18}}
\end{align*}$
