= Solution
Measure $x$ from the static equilibrium position, so the constant gravitational <force> has already cancelled the equilibrium <spring> <force>. With stationary oil, <Newton's second law> gives $m\ddot x=-b\dot x-kx$. The characteristic exponents are $-\gamma\pm i\omega_1$, where
$$
\boxed{\gamma=\frac{b}{2m},\qquad\omega_1=\sqrt{\frac{k}{m}-\frac{b^2}{4m^2}},\qquad b^2<4mk.}
$$
For the stipulated <phase> of the damped cosine,
$$
U(t)=\frac12kA^2e^{-2\gamma t}\cos^2(\omega_1t),\qquad
K(t)=\frac12mA^2e^{-2\gamma t}[\gamma\cos(\omega_1t)+\omega_1\sin(\omega_1t)]^2.
$$
At $t_n=2\pi n/\omega_1$, the <velocity> is not zero: $\dot x=-\gamma Ae^{-\gamma t_n}$. Thus
$$
E_n=\tfrac12(k+m\gamma^2)A^2e^{-2\gamma t_n},\qquad
\boxed{\frac{E_{n+1}}{E_n}=e^{-4\pi\gamma/\omega_1},\quad
\frac{E_n-E_{n+1}}{E_n}=1-e^{-4\pi\gamma/\omega_1}.}
$$
Expanding $e^{-z}$ at small $z=4\pi\gamma/\omega_1$ gives the requested light-damping loss $z$. The exact instantaneous <energy> also satisfies $dE/dt=-b\dot x^2\leq0$.
From the static extension, $k=mg/(0.20\,\mathrm m)=44.145\,\mathrm{N\,m^{-1}}$; the instantaneous drag measurement gives $b=0.10\,\mathrm{N\,s\,m^{-1}}$. Consequently $\gamma=0.05556\,\mathrm{s^{-1}}$ and $\omega_1=7.003\,\mathrm{s^{-1}}$. Two successive downward equilibrium crossings are one damped period apart, so the same <energy> ratio applies. At the first crossing $E=mv^2/2=1.125\times10^{-3}\,\mathrm J$. Therefore
$$
\boxed{\Delta E=1.07\times10^{-4}\,\mathrm J,\qquad
v_{\rm next}=v e^{-2\pi\gamma/\omega_1}=0.04757\,\mathrm{m\,s^{-1}}.}
$$
The <speed> has \b[decreased by $2.43\times10^{-3}\,\mathrm{m\,s^{-1}}=0.243\,\mathrm{cm\,s^{-1}}$]. These values use the exact decay factor, not the first-order approximation.
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