The rate of increase of the population of a city is proportional to the population present at that instant. In the period of 40 years the population increased from 30,000 to 40,000 . At any time $t$ the population is $(a)(b)^{\frac{t}{40}}$. Then the value of $a$ and $b$ are respectively

The population growth follows the differential equation $\frac{dP}{dt} = kP$, leading to the solution $P(t) = P_0 e^{kt}$ with $P_0 = 30,000$.

Using the condition $P(40) = 40,000$, we get $40,000 = 30,000 e^{40k}$, so $e^{40k} = \frac{4}{3}$.

Rewriting $P(t) = 30,000 e^{kt}$ as $P(t) = 30,000 (e^{40k})^{\frac{t}{40}}$, then $P(t) = 30,000 \left(\frac{4}{3}\right)^{\frac{t}{40}}$.

This matches the form $(a)(b)^{\frac{t}{40}}$ with $a = 30,000$ and $b = \frac{4}{3}$, corresponding to option B.

Final answer: $\boxed{\text{B}}$