Here's the question again: what is the expected value of $Z^2$ when $Z\backsim \frac{n+1}{n}Y$ and $\displaystyle Y\backsim f(y\mid\theta )=\frac{n}{\theta}\left(\frac y \theta \right)^{n-1}$ from $0$ to $\theta$
I know the set up of the problem is to make it $$\int_0^\theta z^2f(z\mid \theta) \, dz $$ but the problem is how to figure out the density for z is. I also do not know what the variance is, so I cannot use that formula to figure that out.
$$ \operatorname{E} \left( \left( \frac {n+1} n Y \right)^2 \right) = \left( \frac {n+1} n \right)^2 \operatorname{E}(Y^2). $$ So you need to find $\operatorname{E}(Y^2)$.
\begin{align} \operatorname{E}(Y^2) & = \int_0^\theta y^2 \Big( \frac n \theta \left( \frac y \theta\right)^{n-1} \, dy\Big) = \theta^2 \int_0^\theta \left( \frac y \theta\right)^2 n\left( \frac y \theta \right)^{n-1} \frac{dy} \theta \\[10pt] & = \theta^2 \int_0^1 u^2\cdot nu^{n-1} \, du = \theta^2 \frac n {n+2}. \end{align}
PS: If you're ultimately looking for $\operatorname{var}\left( \dfrac{n+1} n Y \right)$, you will have \begin{align} & \operatorname{var}\left( \frac {n+1} n Y \right) = \left( \frac {n+1} n \right)^2 \operatorname{var}(Y) = \left( \frac {n+1} n \right)^2 \Big( \operatorname{E} (Y^2) - \left( \operatorname{E}(Y) \right)^2 \Big) \\[10pt] = {} & \left( \frac{n+1} n \right)^2 \left( \theta^2 \left( \frac n {n+2} \right) - \theta^2 \left(\frac n {n+1}\right)^2 \right) = \cdots \end{align} ok, so far --- what is just above is routine. The next part is trivial and thus you may think it doesn't need much attention, but the answer is somewhat enlightening: If I'm not mistaken, it comes down to this: $$ \frac {\theta^2}{n(n+2)}. $$