Answer
$3\times10^{5}Pa$
Work Step by Step
Here we use equations 15.2 $W=P\Delta V$ and 15.1 $Q=\Delta U+W$ to find the pressure.
$W=P\Delta V=>P=\frac{W}{\Delta V}-(1)$
$Q=\Delta U+W=>W=Q-\Delta U-(2)$
(2)=>(1),
$$P=\frac{Q-\Delta U}{\Delta V}$$
Let's plug known values into this equation.
$P=\frac{1500\space J-(+4500\space J)}{-0.01\space m^{3}}=3\times10^{5}Pa$ (Pressure of the system)
