:)

$\begin{aligned} S & = \lim_{n \to \infty} \sum_{k=1}^n 2^{2n} \sin^4 \frac \pi {2^n} \\ & = \lim_{n \to \infty} \sum_{k=1}^n 2^{2n} \sin^2 \frac \pi {2^n} \cdot \sin^2 \frac \pi {2^n} \\ & = \lim_{n \to \infty} \sum_{k=1}^n 2^{2n} \sin^2 \frac \pi {2^n} \left(1 - \cos^2 \frac \pi {2^n} \right) \\ & = \lim_{n \to \infty} \sum_{k=1}^n 2^{2n} \left(\sin^2 \frac \pi {2^n} - \sin^2 \frac \pi {2^n} \cos^2 \frac \pi {2^n} \right) \\ & = \lim_{n \to \infty} \sum_{k=1}^n 2^{2n} \left(\sin^2 \frac \pi {2^n} - \frac 14 \sin^2 \frac \pi {2^{n-1}} \right) \\ & = \lim_{n \to \infty} \left(\sum_{k=1}^n 2^{2n} \sin^2 \frac \pi {2^n} - \sum_{k=0}^{n-1} 2^{2n} \sin^2 \frac \pi {2^n} \right) \\ & = \lim_{n \to \infty} 2^{2n} \sin^2 \frac \pi {2^n} = \lim_{n \to \infty} \frac {\sin^2 \frac \pi {2^n}}{\frac 1{2^{2n}}} & \small \blue{\text{Let }x = \frac \pi {2^n}} \\ & = \lim_{x \to 0} \frac {\pi^2 \sin^2 x}{x^2} = \pi^2 \approx \boxed{9.870} & \small \blue{\text{Since }\lim_{x \to 0}\frac {\sin x}x = 1} \end{aligned}$

Hint: Consider a general Problem $\displaystyle\sum_{n=1}^{\infty} 2^{2n}\sin^4\frac{a}{2^n}$ and break $\sin^4x=\sin^2x\sin^2x$