The volume of the block is calculated as $V = l \times b \times t = 12\text{ cm} \times 6\text{ cm} \times 2.45\text{ cm} = 176.4\text{ cm}^3$. According to the rules for significant figures in multiplication, the final result should retain as many significant figures as there are in the original measurement with the least significant figures. Here, the breadth $b = 6\text{ cm}$ has the least number of significant figures, which is $1$. Therefore, the calculated volume must be rounded off to $1$ significant figure. $176.4\text{ cm}^3 = 1.764 \times 10^2\text{ cm}^3$. Rounding this to $1$ significant figure gives $2 \times 10^2\text{ cm}^3$.