Given that 
Initial concentration, [R1] = 0.03
Final concentration, [R2] = 0.02
Time taken ∆t = 25 min = 25 × 6 0 = 1500 sec (1 min = 60 sec )
The formula of average rate of change 

Given that 
Initial concentration [A₁] =0.5
Final concentration [A₂] =0.4
Time is  = 10 min

Rate of reaction = Rate of disappearance of A.
Rate of reaction = $-\frac{1}{2}\frac{∆\left[\mathrm{A}\right]}{∆\mathrm{t}}$
                         $$\begin{gathered} = - \frac{1}{2}\frac{(0.4-0.5) \mathrm{mol} {\mathrm{L}}^{-1}}{10 \mathrm{minute}} \\ = \frac{0.1 \mathrm{mol} {\mathrm{L}}^{-1}}{5 \mathrm{minutes}} \\ = 0.005 \mathrm{mol} {\mathrm{litre}}^{-1} {\min }^{-1}. \end{gathered}$$

The order of the reaction is sum of the powers on concentration.
So that sum will 

r = k[A]^1/2[B]²

Order of reaction = $\frac{1}{2}+2 = 2.5.$

This reaction follows second order kinetics.
So that, the rate equation for this reaction will 
Rate, R = k[X]² .............(1)
Let initial concentration is x mol L⁻¹,
Plug the value in equation (1)
Rate, R₁ = k .(a)²
= ka²
Given that concentration is increasing by 3 times so new concentration will 3a mol L⁻¹
Plug the value in equation (1) we get 
Rate, R₂ = k (3a)²
= 9ka²
We have already get that R₁ = ka₂ plus this value we get
R₂ = 9 R₁ 
So that, the rate of formation will increase by 9 times.
Rate = k[A]²
If concentration of X is increased to three times,
Rate = k[3A]²
or Rate = 9 k A²
Thus, rate will increase 9 times.