The rate of a certain biochemical reaction at physiological temperature (T) occurs $$10^6$$ times faster with enzyme than without. The change in the activation energy upon adding enzyme is:

For any reaction, the Arrhenius equation relates the rate constant $$k$$ to the activation energy $$E_a$$ as

$$k = A\,e^{-E_a/RT}$$

where $$A$$ is the pre-exponential factor, $$R$$ is the gas constant, and $$T$$ is the absolute temperature. We are told that in the presence of the enzyme the reaction is $$10^6$$ times faster, that is

$$\dfrac{k_{\text{enzyme}}}{k_{\text{no enzyme}}}=10^6.$$

Using the Arrhenius form for each rate constant, we have

$$\dfrac{A\,e^{-E_{a,\text{enzyme}}/RT}}{A\,e^{-E_{a,\text{no enzyme}}/RT}} = 10^6.$$

The pre-exponential factor $$A$$ cancels, giving

$$e^{-\dfrac{E_{a,\text{enzyme}}}{RT}} \, e^{+\dfrac{E_{a,\text{no enzyme}}}{RT}} = 10^6.$$

Combining the exponents,

$$e^{-\dfrac{E_{a,\text{enzyme}} - E_{a,\text{no enzyme}}}{RT}} = 10^6.$$

Taking natural logarithms on both sides, we get

$$-\dfrac{E_{a,\text{enzyme}} - E_{a,\text{no enzyme}}}{RT} = \ln(10^6).$$

Now $$\ln(10^6) = 6\ln(10) = 6(2.303).$$ Substituting this value,

$$-\dfrac{E_{a,\text{enzyme}} - E_{a,\text{no enzyme}}}{RT} = 6(2.303).$$

Multiplying both sides by $$-RT$$ gives the change in activation energy:

$$E_{a,\text{enzyme}} - E_{a,\text{no enzyme}} = -6(2.303)RT.$$

Thus the activation energy decreases by $$6(2.303)RT$$ when the enzyme is added.

Hence, the correct answer is Option A.