Area in three significant figures:
$A = 6.08^2 = 37.0 ~ \text{m}^2$
$A_\text{max} = (6.08 + 0.01)^2 = 37.1 ~ \text{m}^2$
$A_\text{min} = (6.08 - 0.01)^2 = 36.9 ~ \text{m}^2$
Hence,
$A = 37.0 \pm 0.10 ~ \text{m}^2$
$\text{Percentage uncertainty} = \dfrac{0.10}{37.0} \times 100\%$
$\text{Percentage uncertainty} = 0.27%$ ← answer
