Compare optical Sensitivity of two sensors
I'm going to assume that are able to get the readings from each sensor and that they are operational.

You will need to find a way of controlling the light into the sensor in a controllable way.  This can be either through ND filters (ND = Neutral Density) a controlled aperture or through chopping.  A controlled aperture can be a lens with a controlled F/# 1, stop is typically 2x the amount of light.  A chopper can be a spinning disk with 1/2 or 1/4 of the disks covered/open and it needs to spin fast enough that the sensor cannot keep up, it sees the "average"

If you reduce the ratio of light by 1/2 and even better also by 1/4 and compare the readings, it will be obvious which one has the higher gain.

From the comment thread : "some more info Device A has 439counts/uW/cm2 and device B has 3 counts/uW.cm2. 22 μW/cm2 is the irradiance for White LED source"

439/3 => 146.33:1 increaased sensitivity.

Let's review what is happening.  The dynamic range is the ratio of the maximum signal to inherent noise of the sensor.  When you increase the sensitivity you are actually increasing the gain of the circuit this amplifies the gain of the maximum and also the noise.

Let's have a made up scenario to illustrate:
sensor X is low sensitivity and requires 2000 w/cm^2 at maximum.  If you increase the light the output saturates (also known as clipping).  the noise level equivalent to 4 w/cm^2.  the sensor is nicely calibrated so the the digital number is 2000 in the bright and 4 in the dark.

The DR = 2000/4 = 500:1

I increase the sensitivity by 100 X, that means that sensor will put out it's maximum of 2000 at 20 w/cm^2 and the noise level will be 4*100 = 400. The sensor with the higher sensitivity will by necessity have a worse DR in digital outputs 2000/400 = 5:1 in terms of light levels, 20/4 = 5:1.