Answer
$\color{blue}{6}$
Work Step by Step
$\begin{align*}
E(Y) &= \int_{\mathbb{R}} y\cdot f_Y(y)\ dy \qquad [\ \text{Definition of}\ E(Y)\ ] \\
&= \int_0^k y\cdot\left(\frac{2y}{k^2}\right)\ dy \qquad \biggl[\ \text{since}\ f_Y(y) = \frac{2y}{k^2},\ y\in[0,k]\ \biggr] \\
&= \frac{2}{k^2}\int_0^k y^2\ dy \\
&= \frac{2}{k^2}\left( \frac{y^3}{3}\ \right\vert_0^k \\
&= \frac{2}{k^2}\left( \frac{k^3}{3} - \frac{0^3}{3} \right) \\
&= \frac{2}{k^2}\left(\frac{k^3}{3} -0 \right) \\
E(Y) &= \frac{2k}{3} \\
\mu &= \frac{2k}{3}
\end{align*}$
$\begin{align*}
E(Y^2) &= \int_{\mathbb{R}} y^2\cdot f_Y(y)\ dy \qquad [\ \text{Definition of}\ E(Y^2)\ ] \\
&= \int_0^k y^2\cdot\left(\frac{2y}{k^2}\right)\ dy \qquad \biggl[\ \text{since}\ f_Y(y) = \frac{2y}{k^2},\ y\in[0,k]\ \biggr] \\
&= \frac{2}{k^2}\int_0^k y^3\ dy \\
&= \frac{2}{k^2}\left( \frac{y^4}{4}\ \right\vert_0^k \\
&= \frac{2}{k^2}\left( \frac{k^4}{4} - \frac{0^4}{4} \right) \\
&= \frac{2}{k^2}\left(\frac{k^4}{4} -0 \right) \\
E(Y^2) &= \frac{k^2}{2}
\end{align*}$
By Theorem 3.6.1, since $\mu= \dfrac{2k}{3}$ exists and $E(Y^2) = \dfrac{k^2}{2}$ is finite for any real number $k$,
$\begin{align*}
\text{Var}(Y) &= E(Y^2) - \mu^2 \\
&= \frac{k^2}{2} - \left(\frac{2k}{3}\right)^2 \\
&= \frac{k^2}{2} - \frac{4k^2}{9} \\
&= \frac{9k^2-8k^2}{18} \\
\text{Var}(Y) &= \frac{k^2}{18}
\end{align*}$
Thus, if $\text{Var}(Y) = 2$, then
$\begin{align*}
\frac{k^2}{18} &= 2 \\
k^2 &= 36 \\
k &= \pm 6 \\
\color{blue}{k}\ &\color{blue}{= 6,}
\end{align*}$
since $k\gt 0$ as $f_Y(y) = \dfrac{2y}{k^2},\ 0 \le y \le k.$
