For a first order reaction A $$\to$$ B

$$x$$ = _________ min. (Nearest integer)

For a first-order reaction the integrated rate law is
$$\ln\!\left(\frac{[A]_0}{[A]}\right)=kt$$

Here $$[A]_0=0.6500\text{ M}$$ at $$t=0$$.
We first evaluate the rate constant $$k$$ using the data at $$t=20\ \text{min}$$,
$$[A]=0.00065\text{ M}$$.

$$\ln\!\left(\frac{0.6500}{0.00065}\right)=k(20)$$

Since
$$\dfrac{0.6500}{0.00065}=1000$$, and
$$\ln(1000)=\ln(10^3)=3\ln10=3(2.302585)=6.907755$$,
we get
$$k=\frac{6.907755}{20}=0.3454\ \text{min}^{-1}$$

Next, let $$x$$ be the time (in minutes) when $$[A]=0.0650\text{ M}$$.
Applying the same rate law:

$$\ln\!\left(\frac{0.6500}{0.0650}\right)=k\,x$$

The ratio is $$\dfrac{0.6500}{0.0650}=10$$, hence $$\ln10=2.302585$$.
Therefore
$$x=\frac{\ln10}{k}=\frac{2.302585}{0.3454}=6.67\ \text{min}$$

Rounding to the nearest whole number gives $$x\approx7\ \text{minutes}$$.

Final Answer: 7