If we take Z-transform of a signal & find its ROC. How to use this ROC? There are ∞ signals in ROC, suppose we choose any signal what to do with it?
Of particular interest is the unit circle ($z=e^{j\omega}$) when it falls within the region of convergence, since the Z-transform $H(z)$ with $z=e^{j\omega}$ is the Fourier Transform. If we compute the Z-Transform of the system’s impulse response (unit sample response), this will give us the frequency response of the system. We can thus use the ROC to determine if the Fourier Transform even exists.
The Z-transform of a discrete function is only unique if the ROC is also provided. Thus, and importantly, by knowing the ROC we can uniquely determine the original discrete function by using the inverse Z-transform. If the system is known to be causal then the ROC is known to be the outside of a circle which resides on the inner-most pole in the z-plane. This property of the relationship of the ROC allows us to mathematically work with non-causal signals by specifying the ROC as a ring.
The actual locations of the singularities (poles and zeros) provides a lot of additional information about the system not provided with the Fourier Transform alone, and these singularities would be the primary locations of interest within the ROC.
Here is another perhaps more intuitive view that expands on the unit circle of the Z-transform being the Fourier Transform and the concept of convergence with regards to the Z-transform result: Every circle centered on the origin within the ROC is the result of the Fourier Transform of the original function multiplied by a real decaying or growing function given by $|z|^{-n}$ where $n$ is a sample number. Thus if the unit circle is not in the ROC, we can determine from ROC a weighting function that we can use in order to determine the Fourier Transform. We see this by comparing the formulas for the z-transform with the Discrete-Time Fourier Transform (DTFT) for an arbitrary function $x[n]$:
DTFT
$$X(\omega) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}$$
Z-Transform
$$X(z) = \sum_{n=0}^{\infty}x[n]z^{-n}$$
$z$ is a complex variable that has a magnitude and phase given as $|z|e^{j\omega}$ (in general the form $Ae^{j\phi}$ is $A\angle \phi$ for real $A$ and real angle $\phi$), and thus the z-transform is related to the DTFT as:
$$X(z) = \sum_{n=0}^{\infty}x[n]z^{-n} = \sum_{n=-\infty}^{\infty}u(t)x[n]|z|^{-n}e^{-j\omega n}$$
Where $u(t)$ is the unit step function.
Where we see that the z-transform is the Discrete Time Fourier Transform of the causal function $u(t)x[n]$ weighted by the real function $|z|^{-n}$ which is  decaying or growing with $n$ (or constant if on the unit circle).
Following an example provided by the OP in the comments:

Consider a signal having Z-transform X(z) = 3+(2/z) &amp; its ROC is whole
z-plane except 0. Suppose z=1, so X(z) =5

With the explanations above we can see intuitively what is occurring and what this means. First, consider how the z-transform on the unit circle is a correlation of our original function $x[n]$ with the set of complex rotating phasors (individual frequency tones) given by $z^n = e^{j\omega n}$, given that correlation is simply a sum of complex conjugate products. When $z=1$ we are correlating our function $x(t)$ with $z^{n}= 1$ for all $n$, basically a DC signal. So $5$ is the DC response of our system, assuming $x[n]$ was the impulse response (unit sample response) of the system, as then the FT is the frequency response and this would imply that a DC signal at the input to the system would have a gain of $5$. If $x[n]$ wasn't an impulse response but an arbitrary signal, $X(z)=5$ for $z=1$ would imply that the DC content of the signal $x[n]$ (which is the mean value) would be $5$ in proportion to the levels everywhere else for $z$ on the unit circle (basically the FT at $f=0$ or DC, which is a correlation of our signal to all possible frequencies).