Q:

Low Pass FIR Filter

I want to design FIR low-pass filter with the following specifications: 
 Cut-off frequency: 200 Hz 

  Order: 20 

  Sampling frequency: 1000 Hz 

what should be the stepwise method to do so 

A:

Your filter is highly under-specified, so I assume that your design is supposed to be very basic. A very basic method would be to simply truncate and shift the impulse response of an ideal low pass filter with cutoff frequency $\omega_c=2\pi f_c/f_s=0.4\pi$ (where $f_s$ is the sampling frequency):
$$h_{ideal}(n)=\frac{\sin(\omega_c n)}{\pi n}$$
Note that since your filter must be causal you need to shift and truncate the ideal impulse response such that it is symmetric with respect to its maximum:
$$h(n)=\frac{\sin(\omega_c (n-10))}{\pi (n-10)},\quad n=0,1,\ldots,20\tag{1}$$
Equation (1) gives you the 21 filter coefficients $h(n)$ of a causal FIR filter approximating an ideal low pass filter response. Note that the filter order is 20.
A simple Matlab/Octave code could look like this:
n = -10:10;
omc = 0.4*pi;           % normalized cut-off frequency in rad
h = sin(omc*n)./(pi*n); % impulse response
h(11) = omc/pi;         % correct NaN value at n=0
H = fft(h,1024);        % complex frequency response
f = 1000/1024*(0:512);  % FFT frequency grid up to fs/2
plot(f,abs(H(1:513)));  % plot magnitude of frequency response

