$a=0.2,~b=0.1$
$\dfrac {dV}{dt} = a V e^{-bt}$
$\dfrac{dV}{V} = a e^{-b t} ~dt$
$\ln(V) = -\dfrac a b e^{-b t} + C$
$V(t) = e^{-\frac a b e^{-b t}}e^C = e^C e^{-2e^{-0.1 t}}$
$V(0) = e^C e^{-2} = 1.86$
$e^C \approx 13.7436$
$V(t) \approx 13.7436 e^{-2e^{-0.1 t}}$
Another approach is to write the given ODE in standard linear form:
$\displaystyle \frac{dV}{dt}-0.2e^{-0.1t}V=0$
Compute the integrating factor:
$\displaystyle \mu(t)=\exp\left(-0.2\int e^{-0.1t}\,dt\right)=e^{2e^{-0.1t}}$
Multiply through by the integrating factor:
$\displaystyle e^{2e^{-0.1t}}\frac{dV}{dt}-0.2e^{-0.1t}e^{2e^{-0.1t}}V=0$
Observe that the LHS can be rewritten as the derivative of a product:
$\displaystyle \frac{d}{dt}\left(e^{2e^{-0.1t}}V\right)=0$
Integrate through with respect to $\displaystyle t$:
$\displaystyle e^{2e^{-0.1t}}V=C$
Thus:
$\displaystyle V(t)=Ce^{-2e^{-0.1t}}$
Use the initial conditions to find $\displaystyle C$
$\displaystyle V(0)=Ce^{-2e^{-0.1(0)}}=Ce^{-2}=1.86\implies C=1.86e^2$
And so the solution to the IVP is:
$\displaystyle V(t)=1.86e^2e^{-2e^{-0.1t}}=1.86e^{2-2e^{-0.1t}}$
This is equivalent to the solution posted by romsek, just written in a slightly different form.