The inactivation rate of a viral preparation is proportional to the amount of virus. In the first minute…

As the unit of rate constant is ${\min }^{-1}$ so it must be a first order reaction
$\mathrm{k}\times \mathrm{t}=2.303{\mathrm{logA}}_0/{\mathrm{A}}_{\mathrm{t}}$
in $1$ min $10\%$ is inactivated so taking
${\mathrm{A}}_0=100 { \mathrm{A}}_{\mathrm{t}}=90$ in $1 \min$
So $\mathrm{k}\times 1=2.303\times \log \frac{100}{90}$

$=2.303\times (\log 10-2\log 3)$
$=2.303\times (1-2\times 0.477)$
$=0.10593$

$=105.93\times {10}^{-3}$
Answer is $106$