I also find the question puzzling, for reasons that will be described in the comment at the end. But the following is certainly true.
Let $z_a$ be a particular solution of the equation that has $F_0=a\ne 0$.
Let $z_b=(b/a)z_a$. Then $z_b$ is a particular solution of the equation that has $F_0=b$.
Verification is straightforward. Substitute $(b/a)z_a$ for $z_b$ into the left-hand side.
In general the derivative of $kf(t)$, where $k$ is a constant, is kf'(t), and the second derivative is kf''(t). Thus $\frac{d^2z_b}{dt^2}+\Gamma\frac{dz_b}{dt}+\omega_0^2z_b=(b/a)\left(\frac{d^2z_a}{dt^2}+\Gamma\frac{dz_a}{dt}+\omega_0^2z_a\right).$
Note that linearity was essential. The argument could not be pushed through with a nonlinear equation like $\frac{d^2y}{dt^2}+y\frac{dy}{dt}+y=F(t)$.
Comment: The wording of the problem is imprecise. We have shown that given a particular solution $z_1$ for $F_0=1$, we can find a particular solution $z_c$ for $F_0=c$ by just multiplying $z_1$ by $c$. But not all particular solutions arise in this way.
I will illustrate this with the simpler equation $\frac{dy}{dt} +y =5.$
One particular solution is $y=5$ (the constant function $5$).
Now look at the equation $\frac{dy}{dt}+y=105.$
We can certainly get a particular solution of this equation by multiplying $5$ by $(105/5)$.
But the equation $\frac{dy}{dt}+y=105$ has plenty of other particular solutions, like $y=e^{-t}+105$. This particular solution cannot be obtained by multiplying the particular solution $5$ by a proportionality constant!
Note that the ratio $z_b/z_a$ is exactly $b/a$, which is the ratio of two values of $F_0$. This gives the required proportionality.