Answer
$\color{blue}{\frac{3}{2}\ \text{in}^3 = 1.5\ \text{in}^3}$
Work Step by Step
For each randomly chosen $ Y \in (0,1),$ the volume $V$ of the box to be constructed is equal to the area of the square base, which is $Y^2$, multiplied by the box's height of five (5) inches. Thus, $V = 5Y^2$.
Computing for the expected volume of the box (in cubic inches), we have
\begin{align*}
E(V) &= E(5Y^2) \qquad \text{since}\ V = 5Y^2 \\
&= \int_{\mathbb{R}} 5y^2\cdot f_Y(y)\ dy \qquad [\ \text{Definition of}\ E(Y^2)\ ]\\
&= \int_0^1 5y^2\cdot(6y(1-y))\ dy \qquad [\ \text{since}\ f_Y(y)=6y(1-y),\ y\in[0,1]\ ] \\
&= 30\int_0^1(y^3-y^4)\ dy \\
&= 30\left( \frac{y^4}{4} - \frac{y^5}{5}\right)_0^1 \\
&= 30\left[\left( \frac{1^4}{4} - \frac{1^5}{5}\right) - \left( \frac{0^4}{4} - \frac{0^5}{5}\right)\right] \\
&= 30\left[\left(\frac{1}{4}-\frac{1}{5}\right) - (0-0)\right] \\
&= 30\left(\frac{1}{20}\right) \\
\color{blue}{E(V)}\ &\color{blue}{= \frac{3}{2}\ \text{in}^3 \;=\; 1.5\ \text{in}^3}
\end{align*}
