I'm going to assume that your function $f(r):=\sin^4 (\pi r/R)\ (r/R)$ is the radial expression of the radial function of two real variables $f(x,y):= \sin^4 (\pi \sqrt{x^2+y^2}/R)\ (\sqrt{x^2+y^2}/R)$ defined in the disc $D:=\{ (x,y)\in \mathbb{R}^2:\ x^2+y^2\leq R^2\}$, because otherwise it will have no meaning to average $f(r)$ over an area.
The average of such an $f(x,y)$ over $D$ is then given by:
$$\overline{f} := \frac{1}{\operatorname{area}(D)}\ \iint_D f(x,y)\ \text{d}x\text{d}y\; ;$$
the integral in the RH side can be computed using polar coordinates, hence:
$$\begin{split}
\overline{f} &:= \frac{1}{\pi R^2}\ \int_0^R \int_0^{2\pi} f(r)\ r\ \text{d}r\text{d}\theta \\
&= \frac{2}{R^2}\ \int_0^R f(r)\ r\ \text{d}r\\
&= \frac{2}{R^2}\ \int_0^R \sin^4 \left(\pi \frac{r}{R}\right)\ \frac{r^2}{R}\ \text{d}r\\
&\stackrel{t=\pi r/R}{=} \frac{2}{\pi R^2}\ \int_0^\pi t^2\ \sin^4 t\ \text{d}t\; .
\end{split}$$
The latter integral can be computed using standard integration by parts techniques, which (after lenghty and tedious computations) yield:
$$\begin{split}
\int_0^\pi t^2\ \sin^4 t\ \text{d}t &= \frac{1}{256} \left( 32 t^3+ 32(2t^2-1)\sin 2t +(8t^2-1)\sin 4t -64t\cos 2t +4t\cos 4t\right)\Bigg|_0^{\pi}\\
&= \frac{\pi}{64}\ (8\pi^2-15)\; ,
\end{split}$$
therefore:
$$\overline{f} = \frac{8\pi^2-15}{32R^2}\; .$$