I would solve for $t$
$S(t)=\displaystyle\sum_{n=1}^{t}\dfrac{1}{1.08^{n}}=3.$
The sum of a geometric progression with first term $u_{1}$ and ratio $r$ is given $(\ast)$ by
$S(t)=u_{1}\times \dfrac{1-r^{t}}{1-r}.$
In this case
$\dfrac{1}{1.08}\times \dfrac{1-\left( \dfrac{1}{1.08}\right) ^{t}}{1-\dfrac{1}{% 1.08}}=3$
or
$\dfrac{1-\left( \dfrac{1}{1.08}\right) ^{t}}{0.08}=3\iff \left( \dfrac{1}{1.08}% \right) ^{t}=0.76.$
Applying logarithms, we obtain the "time" $t$
$t\log \left( \dfrac{1}{1.08}\right) =\log 0.76,$
$t=\dfrac{\log 0.76}{-\log 1.08}\approx 3.5659,$
which means that we need more than $3$ periods and less than $4$, at an interest rate of $8\%$, coumpounded per period, to get a total of $3$ currency units.
$(\ast)$ Derivation:
$S=u_{1}+u_{2}+u_{3}+\ldots +u_{t}$
$rS=ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{t-1}+ru_{t}$
$u_{k}=ru_{k-1}=u_{1}r^{k-1}$
$S-rS=\left( u_{1}+u_{2}+u_{3}+\ldots +u_{t}\right) -\left( ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{t-1}+ru_{t}\right) $
$(1-r)S=u_{1}-ru_{t}$
$S=\dfrac{u_{1}-ru_{t}}{1-r}=\dfrac{u_{1}-u_{1}r^{t}}{1-r}$