Jerry needs to paint the inside of the garage - the two side walls and the rear wall. The garage is 6 m long. 3 m wide and 2 m high. He can buy the paint in \( 2-\mathrm{L} \) and in \( 500-\mathrm{mL} \) tins. One L willcover \( 14 \mathrm{~m}^{2} \).

To calculate the total area Jerry needs to paint, we first determine the area of the two side walls and the rear wall. The two side walls are each \( 2 \, \text{m} \times 3 \, \text{m} = 6 \, \text{m}^2 \) (2 walls, so \( 6 \, \text{m}^2 \times 2 = 12 \, \text{m}^2 \)) and the rear wall is \( 6 \, \text{m} \times 2 \, \text{m} = 12 \, \text{m}^2 \). This brings the total area to \( 12 \, \text{m}^2 + 12 \, \text{m}^2 = 24 \, \text{m}^2 \). Now, considering the coverage, Jerry will need \( \frac{24 \, \text{m}^2}{14 \, \text{m}^2/\text{L}} \approx 1.71 \, \text{L} \) of paint. Since he can buy in \( 2 \, \text{L} \) or \( 500 \, \text{mL} \) tins, a single \( 2 \, \text{L} \) tin will suffice. For an extra tip, always buy a little extra paint to account for touch-ups or any mistakes! It’s a great idea to keep leftover paint for future projects too.