For a first order reaction, $$(A) \rightarrow$$ products, the concentration of $$A$$ changes from $$0.1$$ M to $$0.025$$ M in $$40$$ minutes. The rate of reaction when the concentration of $$A$$ is $$0.01$$ M is :

The integrated rate law for a first-order reaction $$A \rightarrow$$ products is
$$k = \frac{1}{t}\,\ln\!\left(\frac{[A]_0}{[A]}\right)$$

Initial concentration $$[A]_0 = 0.1\,\text{M}$$, final concentration after $$t = 40\,\text{min}$$ is $$[A] = 0.025\,\text{M}$$. Substituting,

$$k = \frac{1}{40}\,\ln\!\left(\frac{0.1}{0.025}\right) = \frac{1}{40}\,\ln(4)$$

Since $$\ln(4) = 1.3863$$,

$$k = \frac{1.3863}{40} = 0.03466\,\text{min}^{-1}$$

For a first-order reaction, instantaneous rate at any moment is
$$\text{rate} = k\,[A]$$

At the required concentration $$[A] = 0.01\,\text{M}$$,

$$\text{rate} = 0.03466 \times 0.01 = 3.466 \times 10^{-4}\,\text{M min}^{-1}$$

Rounded to three significant figures, $$\text{rate} = 3.47 \times 10^{-4}\,\text{M min}^{-1}$$

Option B which is: $$3.47 \times 10^{-4}$$ M/min