As Michal Podmanický pointed out, your capacitor is going to take thousands of seconds (hours) to charge, and they showed this from an energy perspective. That's fine, but it doesn't account for the gradually reducing charge current over time, assuming instead that the power being delivered to the capacitor is constant.
Your formula for the time it will take to charge to 5.4V via resistance \$R\$ is close, but not correct. The formula for the time it takes for a capacitor's voltage to go from some zero to some value \$V\$, on its way towards asymptote \$V_A\$, is:
$$
\begin{aligned}
t &= -RC \cdot ln\left(\frac{V_A - V}{V_A}\right) \\ \\
&= -RC \cdot ln\left(1-\frac{V}{V_A}\right) \\ \\
\end{aligned}
$$
I think you swapped \$V\$ and \$V_A\$. Anyway, this gives us:
$$
\begin{aligned}
t &= -R \times 0.47 \times ln\left(1-\frac{5.4}{5.5}\right) \\ \\
&= R \times -0.47 \times -4.0 \\ \\
&= 1.9R
\end{aligned}
$$
You must choose R to pass whatever current the panel can provide at about 5V, which seems to be about 30μA. This is initial current, current that will flow when the capacitor has 0V across it, and the remaining 5V is across resistance R. By Ohm's law:
$$ R = \frac{V}{I} = \frac{5V}{30\mu A} = 167k\Omega $$
Plugging that value for R into the time equation:
$$
\begin{aligned}
t &= 1.9R \\ \\
&= 1.9 \times 167k\Omega \\ \\
&= 317,000s \\ \\
&= 88 hr
\end{aligned}
$$
Sorry to disappoint. On top of that, due to the exponential nature of the charge/discharge curves (using a fixed resistance), charging is extremely slow near the end of charging. That means capacitor voltage will fall much more quickly in the early stages of discharge than it will rise when power returns.
It's not all bad news though. You don't have to use a fixed resistance, and assuming the load draws a similar current, once the capacitor is charged it will power that load for days.
Here's a design that will charge the capacitor with about 10μA constant current, right up until full charge (as opposed to ever-diminishing current as the capacitor charges):
simulate this circuit – Schematic created using CircuitLab
With a constant charge current, capacitor voltage rises linearly, instead of exponentially, and charges fully somewhat quicker, even though charge current is significantly less than 30μA. It also means that even when nearly fully charged, charge current doesn't diminish, and the capacitor charges at the same rate, right up until it's full. Here's a plot of capacitor voltage (orange) as input voltage (blue) from the solar panel varies:
Just be aware that whatever current the capacitor is taking (during charging), that current is not available for the load, so with 10μA going to the capacitor, out of the 30μA available, the load can draw a maximum of only 20μA.
