$h=\frac{V_0}{A}=\frac{40}{8}=5\text{\hspace{0.17em}cm}$ 
Let speed of piston $=u$
Continuity equation
 $Au=a{u}_0 \Rightarrow u=\frac{a}{A}{u}_0=\frac{2{\text{\hspace{0.17em}mm}}^{\text{2}}}{8{\text{\hspace{0.17em}cm}}^{\text{2}}}\times 2\text{\hspace{0.17em}m/s\hspace{0.17em}=\hspace{0.17em}0.005\hspace{0.17em}m/s}$
Applying Bernoulli’s equation between point 1 and 2-
$P_0+\frac{1}{2}{\text{ρu}}_{\text{0}}^{\text{2}}+\text{\hspace{0.17em}ρ}gh=P+\frac{1}{2}\text{ρ}{u}^2$ 
$\Rightarrow P=P_0+\frac{1}{2}{\text{ρu}}_{\text{0}}^{\text{2}}-\frac{1}{2}{P}_0+\frac{1}{2}\text{ρ}{u}^2+\text{\hspace{0.17em}\hspace{0.17em}ρ}gh$
For piston to go up at constant speed
$P_{0}A+F=PA+Mg\Rightarrow F=\frac{1}{2}\text{ρ}A(u_0^2-u^2)+A\rho gh+Mg$ 
As piston moves up, $h$change from 5 cm to zero. So, $F$ keeps decreasing
$\therefore$                Work done by external agent is
$W=\underset{h}{\overset{0}{\int }}Fdh=\frac{1}{2}\text{ρ}A(u_0^2-u^2)h+\frac{1}{2}A\text{ρ}g{h}^2+Mgh$ 
 $=\frac{1}{2}\text{ρ}A{u}_0^{2}h+\frac{1}{2}A\text{ρ}g{h}^2+Mgh[\because u_0^2>>u^2]=0.115J$