The pressure of a gas changes linearly with volume from A to B as...

Given here, $∆ Q = 0$

From first law of thermodynamics, $∆ Q = ∆ U + W$, where, $∆ U$ is change in internal energy and $W$ is work done.

We have, $∆ U = - W$

$W = AreaunderP - Vcurve = ∫ PdV$

Here, volume decreases, hence work done by system will be negative.

$∆ U = - W = 30 \times {10}^{3} \times 150 \times {10}^{- 6}$

$= 4500 \times {10}^{- 3}$

$= 4 . 5J$