a
Work done $= P \Delta V$
$=3 \times 10^5 \times 1600 \times 10^{-6}$
$=480\,J$
Only $10 \%$ of heat is used in work done.
Hence $\Delta Q=4800\,J$
The rest goes in internal energy, which is $90\,\%$ of heat.
Change in internal energy $=0.9 \times 4800=4320\,J$