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(C) The activity of a radioactive sample at time $t$ is given by the formula: $R = R_0 \left( \frac{1}{2} \right)^{n}$,where $n = \frac{t}{T_{1/2}}$ i

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$108$

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(C) The activity of a radioactive sample at time $t$ is given by the formula: $R = R_0 \left( \frac{1}{2} ight)^{n}$,where $n = \frac{t}{T_{1/2}}$ is the number of half-lives.Given that the initial activity is $64$ times the permissible level $(R = 64 R_0)$,we set the final activity to $R_0$ to reach the safe limit.Thus,$64 R_0 \left( \frac{1}{2} ight)^{n} = R_0$.This simplifies to $\left( \frac{1}{2} ight)^{n} = \frac{1}{64}$.Since $64 = 2^6$,we have $\left( \frac{1}{2} ight)^{n} = \left( \frac{1}{2} ight)^{6}$.Therefore,$n = 6$.Given the half-life $T_{1/2} = 18$ days,the total time $t$ is $t = n 	imes T_{1/2} = 6 	imes 18 = 108$ days.Hence,the laboratory will be safe after $108$ days.

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