For a first order reaction with half-life of 150 seconds, the time taken for the concentration of the reactant to fall from $\frac{\mathrm{M}}{10}$ to $\frac{\mathrm{M}}{100}$ will be approximately

$$\begin{gathered} {\mathrm{t}}_{1/2}=150\sec \\ \begin{array}{rl}\mathrm{k}& =\frac{0.693}{{\mathrm{t}}_1/2}\\ & =\frac{0.693}{150\mathrm{ce}}\end{array} \\ \mathrm{For} \mathrm{first} \mathrm{order} \\ -\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{A}}_0}{{\mathrm{A}}_{\mathrm{t}}} \\ \begin{array}{rl}\mathrm{t}& =\frac{2.303}{\mathrm{k}}\log \frac{{\mathrm{A}}_0}{{\mathrm{A}}_{\mathrm{t}}}\\ & =\frac{2.303}{\frac{0.693}{100}}\times \log \left(10\right)\end{array} \\ \begin{array}{rl} & =\frac{2.303\times 150\times 1}{0.693}\\ & =493.4\\ & \simeq 500\mathrm{cec}\end{array} \end{gathered}$$

Therefore, the correct option is (B).