How to determine the window size and weights in Weighted Moving Average (WMA), given desired cut-off frequency?
Your normalized window is given by

$$w(n)=\frac{2}{N+1}\frac{N-n}{N},\quad n=0,1,\ldots,N-1$$

The window satisfies

$$\sum_{n=0}^{N-1}w(n)=1$$

which means that the gain of the corresponding moving average filter is 1 at DC.

For determining the cut-off frequency, we need to compute the frequency response of the window:

$$W(e^{j\theta})=\sum_{n=0}^{N-1}w(n)e^{-jn\theta}$$

After some algebra you get

$$W(e^{j\theta})=\frac{2}{N+1}\frac{1-\frac{N+1}{N}e^{-j\theta}+\frac{1}{N}e^{-j(N+1)\theta}}{(1-e^{-j\theta})^2}\tag{1}$$

Now you need to find the value of $N$ for which the magnitude of (1) at the cut-off frequency $\theta_c=2\pi\frac{3Hz}{50Hz}$ becomes $1/\sqrt{2}$ (-3dB). Since $N$ must be integer you cannot get any desired cut-off frequency, but the given cut-off is approximately achieved by $N=9$, for which $|W(e^{j\theta_c})|=0.698$ (-3.13dB).