How to obtain the Sound Pressure Level from .wav file based on these given parameters? I have a wav file recorded from a hydrophone. I am not quite familiar with signal processing so I really need some advice from you.

I have gathered the information which looks useful and related.

sample rate = 250 000 Hz

Receive sensitivity of hydrophone = -180 db re 1V/μ Pa

Peak-to-peak voltage range = 5.64 V

Pre-amplifier gain = 18 dB (Modified)

Let A be the amplitude saved in the wav file. What is the formula to transform A from amplitude to Sound Pressure Level?

If the given information is not sufficient, please tell me what more do I need?

Any helps are appreciated!
The hydrophone has -180 dB 1V/uPa or -60 dB V/Pa. 1 Pa produces hence 1mV or, other way around, the sensitivity is 1mV/Pa. Peak voltage is 5.64 peak to peak. That's 2.82V peak or 2V RMS. 

A full scale sine wave has 2 V RMS which is 2000 Pa. dBSPL is referenced against 20uPa so 2000 Pa is equivalent to 160 dBSPL. 

The clipping point of your wave file corresponds to 160 dB SPL. The exact relationship between number and dBSPL  depends  on the format of the wave file (fixed vs. floating, number of bits). For a 16 bit fixed point format, the clipping point corresponds to the number 32768.