Yes. I calculated wrongly.
So, here is the correct one:

For V1 = 10mv ; V2 = 0V.
Output of U2A = .01 + .045 = .055V
Output of U3A =  0 - .045 = -.045V
So, U2A - U3A = .055 - (- .045) = .1V (gain of 10 at first stage)

Oh. I just looked at the instrumentation amplifier to know how its used for measuring the physical quantities such as temperature, light intensity, etc., with the help of a transduser.
But, havnt even though that its mainly because of its ability to filter common mode noise they are used.

My book also stressed only on amplification but not on CMRR in the instrumentation amplifier topic.
just says in one line " Has high CMRR "

Now, i would like to ask how do i know that my signal is afffected with commmon mode noise? I mean to know the factors that superimpose common mode noise in a signal so that i could avoid it.

regard's
Suresh.