In the process given, internal energy is

$U=a V^3$

$\Rightarrow \quad \frac{f n R T}{2} =a V^3$

where, $f=$ degree of freedom $=3$ (for ideal gas) and $n=$ number of moles $=1$.

$\Rightarrow \quad \frac{3}{2} R T=a V^3$

$\Rightarrow \quad \frac{3}{2} p V=a V^3 \quad[\because p V=R T]$

So, pressure in this process is given by

$p=\frac{2 a}{3} V^2$

Now, work done during the process is

$W=\int \limits_V^{2 V} p d V=\int \limits_V^{2 V} \frac{2 a}{3} V^2 d V$

$=\frac{2 a}{3} \times \frac{1}{3}\left((2 V)^3-V^3\right)$

$=\frac{2 a}{9}\left(7 V^3\right)$

$=\frac{2}{9} \times 7 \times \frac{3}{2} R T \quad\left[\because a V^3=\frac{3}{2} R^2 T\right]$

$=\frac{7}{3} R T$