The inactivation rate of a viral preparation is proportional to the amount of virus. In the first minute after preparation, 10% of the virus is inactivated. The rate constant for viral inactivation is ___ $$\times 10^{-3}$$ min$$^{-1}$$.
(Nearest integer) [Use : ln 10 = 2.303; $$\log_{10} 3 = 0.477$$ property of logarithm: $$\log x^y = y \log x$$]

Since the inactivation rate is proportional to the amount of virus, this is a first-order reaction. For a first-order process, the rate constant $$k$$ is given by:

$$k = \frac{1}{t} \ln\left(\frac{N_0}{N}\right)$$

In the first minute (t = 1 min), 10% of the virus is inactivated, so 90% remains. Thus $$N/N_0 = 0.90$$.

$$k = \frac{1}{1} \ln\left(\frac{1}{0.9}\right) = \ln\left(\frac{10}{9}\right) = \ln 10 - \ln 9 = \ln 10 - 2\ln 3$$

Using the given values: $$\ln 10 = 2.303$$ and $$\log_{10} 3 = 0.477$$, so $$\ln 3 = 0.477 \times 2.303 = 1.09843$$.

$$k = 2.303 - 2 \times 1.09843 = 2.303 - 2.19686 = 0.10614 \text{ min}^{-1}$$

Expressing as $$p \times 10^{-3} \text{ min}^{-1}$$: $$p = 106.14 \approx 106$$.

Therefore, the rate constant is $$\boxed{106} \times 10^{-3} \text{ min}^{-1}$$.