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If $1^{3}+2^{3}+..........+9^{3}=2025 then the value (0.11)^{3}+(0.22)^{3}+......+(0.99^{3})$ is close to

A) 0.2695

B) 0.3696

C) 2.695

D) 3.695

Option C

$(0.11)^{3}+(0.22)^{3}+.......+(0.99)^{3}=(0.11)^{3}(1^{2}+2^{2}+......+9^{2})$

$0.001331\times2025=2.695275= 2.695$

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