1. Set up the kinetics equation

Since the inactivation rate is proportional to the amount of virus present, the process follows first-order kinetics:

$\frac{dN}{dt} = -kN$

On integration,

$N = N_0 e^{-kt}$

or

$\frac{N}{N_0} = e^{-kt}$

2. Use the given information

In the first minute, 10% is inactivated, so 90% remains after 1 minute.

Thus,

$\frac{N}{N_0} = 0.9 \quad \text{at} \quad t=1\,\text{min}$

So,

$0.9 = e^{-k(1)}$

Taking natural logarithm:

$\ln 0.9 = -k$

Hence,

$k = -\ln 0.9 = \ln\left(\frac{10}{9}\right)$

3. Evaluate numerically using the given logarithm values

We use

$\ln\left(\frac{10}{9}\right) = \ln 10 - \ln 9$

Now,

$\ln 9 = \ln(3^2)=2\ln 3$

Given:

$\log_{10} 3 = 0.477$

So,

$\ln 3 = 2.303 \times 0.477$

Therefore,

$\ln 9 = 2(2.303 \times 0.477)$

Compute:

$2.303 \times 0.477 = 1.098531$

Thus,

$\ln 9 = 2 \times 1.098531 = 2.197062$

Now,

$\ln 10 = 2.303$

Hence,

$k = 2.303 - 2.197062 = 0.105938$

So,

$k \approx 0.106\,\text{min}^{-1}$

4. Match with the asked format

The question asks for:

$\underline{\hspace{2cm}} \times 10^{-3}\,\text{min}^{-1}$

Since

$0.106 = 106 \times 10^{-3}$

the required integer is:

$\boxed{106}$

5. Comparison with stored answer

Derived answer = $106$

Stored correct answer = $106$

They agree.