Battery recharging via solar panel

Very partial answer to start. More as/if wanted:

You need a maximum of about 3V AT the battery so a 4V panel would work well and a lower one say 3.5V Vmp (voltage at maximum power) is probably OK.

Maximum capacity of an AAA NiCd is liable to be about 500 mAh and may be less.
0.1 C at 500 mAh = 50 mA.
4V x 50 mA = 200 mW rated PC power.
You can buy say 300 mW 4V panels in 500 quantity for around $US1 in China.

Available solar energy is a longish subject in its own right and well enough covered in other SE answers. Look at www.gaisma.com for an idea of how many equivalent SSH (sunshine-hours) a day you can expect. A useful rule of thumb that is close enough for many areas is 2 SSH a day in winter and about 5 in summer. When dimensioning energy available from batteries you use 2 SSH.

By the time you have moved energy from panel to battery and battery to load you'll be lucky to get 50% of notional panel energy out. Working in mAh can give you a better feel. A say 4V 50 mAA Imp panel will deliver 100 mAh in 2 hours of full sun. You'll get much but not all of that out of the battery - say 60-75 mAh, and then you lose energy if you need to convert to another voltage to drive the load. End to end you can typically assume the equivalent of Vbattery mean x 50 mAh for load use. Note that with NiCd you start out at somewhere above 1.2V/cell when fully charged and end up at 1V or so when almost discharged. A lightly loaded cell will be close to 1.2V across much of its life - so in winter a 200 mW panel may give 120 - 150 mWh of energy from 2 equivalent hours of sun. I understand that this is well above the mean energy level that you need in this application - so a much less than 200 mW panel should be adequate.

Even on ebay you can get acceptable prices for small volumes.

eg this seller is offering panels at prices well below the one you showed.
This panel rated at 6V 100 mA costs about $5.

Your approach is basically correct, but I question some of your numbers:

How does "1 day of reserve" only require 30% additional capacity, rather than 100%?

You've neglected to take into account the storage efficiency of the battery itself (Wh in does not equal Wh out), which is only about 70% for lead-acid.

For your 275 mA × 9 h = 2.5 A-h daily load, you need a 10 A-h battery, which accounts for your 100% functional reserve, plus 100% "battery reserve".

Your charger will need to supply 7 A-h per day, which is the functional capacity (including reserve) divided by the storage efficiency of the battery.

If you have a very good charge controller, it might achieve 90% power efficiency (but possibly much less). 7 A-h × 12 V = 84 W-h nominal. 84 W-h / 0.90 efficiency / 4 h (equivalent insolation) = 24 W panel.