Please help?

Suppose "30" degree means 30°C, i.e. 303 K.

One form of Arrhenius equation: ln(k) = [-Ea/(RT)] + ln(A)
Without catalyst: Ea = Ea₁ and k = k₁.  Without catalyst: Ea = Ea₂ and k = k₂
ln(k₁) = [-Ea₁/(RT)] + ln(A) …… [1]
ln(k₂) = [(-Ea₂/(RT)) + ln(A) …… [2]

[2] - [1]:
ln(k₂) - ln(k₁) = [-Ea₂/(RT)] - [-Ea₂/(RT)]
ln(k₂/k₁) = (Ea₁ - Ea₂)/(RT)
Hence, Ea₁ - Ea₂ = RT ln(k₂/k₁)

Now, R = 8.314 × 10⁻³ kJ/mol, T = 303 K, and k₂/k₁ = 3
Decrease in Ea, Ea₁ - Ea₂ = (8.314 × 10⁻³) × 303 × ln(3) = 2.77 kJ/mol

from the arrhenius equation
.. k = Ao exp(-Ea / RT)
for 2 rxns at the temp with Ao constant
.. k1 = Ao exp(-Ea1 / RT)
.. k2 = Ao exp(-Ea2 / RT)
dividing
.. k1 / k2 = (Ao/Ao) exp(-Ea1/RT) / exp(-Ea2/RT)
.. k1 / k2 = exp( (1/RT) * (Ea2 - Ea1))

and we know the general rate equation is of the form
.. rate = k * [A]^n
so that for 2 rxns with the same starting concentration of A
.. rate1 = k1 * [A]^n
.. rate2 = k2 * [A]^n
dividing
.. rate1 / rate2 = k1 / k2 * ([A]^n / [A]^n)
.. rate1 / rate2 = k1 / k2

subbing
.. rate1 / rate2 = exp( (1/RT) * (Ea2 - Ea1))
ln'ing
.. ln(rate1 / rate2) = (1/RT) * (Ea2 - Ea1)
.. (Ea2 - Ea1) = RT * ln*(rate1 / rate2)

and YOU get to finish. ... hint.. rate2 = 3*rate1