The inactivation rate of a viral preparation is proportional to the...

As the unit of rate constant is $\min^{- 1}$ so it must be a first order reaction
$k \times t = 2 . 303{logA}_{0} / {A}_{t}$
in $1$ min $10$ is inactivated so taking
${A}_{0} = 100{A}_{t} = 90$ in $1min$
So $k \times 1 = 2 . 303 \times log\dfrac{100}{90}$

$= 2 . 303 \times ( \log 10 - 2\log 3 )$
$= 2 . 303 \times ( 1 - 2 \times 0 . 477 )$
$= 0 . 10593$

$= 105 . 93 \times {10}^{- 3}$
Answer is $106$