Question on FIR filter
[EDITED] From left to right, try to find the impulse response. On the bottom branch, delay the unit pulse $d$ and get $d[n-1]$. Add $Ad[n]$. Delay twice and get $d[n-3] +Ad[n-2]$. Add $Bd[n]$. Add $Cd[n-1]$. So the FIR coefficients of the impulse response $h$ will be $[1,A,C,B]$, but in the reverse order (as correctly pointed out by colleagues), because $y[n] = \sum d[k]h[n-k]$. So your system is causal, and $h[0]=B$, $h[1]=C$, $h[2]=A$, $h[3]=1$.

However, at $n=20$, you are in a stabilized region of the signal (all the known samples are equal to $1$, so even if you make a mistake as I did first, the answer will be the same, $1+A+C+B$, as stated before by other colleagues.