As pressure against which expansion is happening is constant, work done can be written as =PΔV
=(3×105)×(1600×10−6)
=480J
As only 10 of heat is used in work done.
Hence ΔQ=0.1W=4800J
The rest goes in increase in internal energy, which is 90 of heat.
Change in internal energy =0.9×4800=4320J