1. Relate splitting time to rate constant

Milk splits when the population of lactobacillus doubles. If bacterial growth is treated as a first-order growth process,

$N = N_0 e^{kt}$

For doubling,

$2N_0 = N_0 e^{kt} \Rightarrow e^{kt}=2 \Rightarrow kt = \ln 2$

So the time required for splitting is

$t = \frac{\ln 2}{k}$

Thus, for the two temperatures:

$k_1 = \frac{\ln 2}{t_1}, \qquad k_2 = \frac{\ln 2}{t_2}$

with

$t_1 = 60\ \text{min at } 300\,\text{K}, \qquad t_2 = 40\ \text{min at } 400\,\text{K}$

Hence,

$\frac{k_2}{k_1} = \frac{t_1}{t_2} = \frac{60}{40} = \frac{3}{2}$

Therefore,

$\ln\left(\frac{k_2}{k_1}\right)=\ln\left(\frac{3}{2}\right)=0.4$

2. Use Arrhenius equation

The Arrhenius relation in two-temperature form is

$\ln\left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

Substitute the values:

$0.4 = \frac{E_a}{8.3}\left(\frac{1}{300}-\frac{1}{400}\right)$

Now,

$\frac{1}{300}-\frac{1}{400} = \frac{4-3}{1200} = \frac{1}{1200}$

So,

$0.4 = \frac{E_a}{8.3}\cdot \frac{1}{1200}$

$E_a = 0.4 \times 8.3 \times 1200$

3. Calculate

First,

$8.3 \times 1200 = 9960$

Then,

$E_a = 0.4 \times 9960 = 3984\ \text{J mol}^{-1}$

Convert to kJ mol$^{-1}$:

$E_a = 3.984\ \text{kJ mol}^{-1}$

So the closest value is

$\boxed{4\ \text{kJ mol}^{-1}}$

More precisely,

$E_a \approx 3.98\ \text{kJ mol}^{-1}$

4. Comparison with stored correct answer

Stored correct answer: $3.98$ to $3.99$

Derived answer: $3.984$

These agree.