A reaction is first order in A and second order in B.

(i) The differential rate equation will be

$-\frac{d[\mathrm{R}]}{d t}=k[\mathrm{~A}][\mathrm{B}]^{2}$

(ii) If the concentration of B is increased three times, then

$-\frac{d[\mathrm{R}]}{d t}=k[\mathrm{~A}][3 \mathrm{~B}]^{2}$

$=9 \cdot k[\mathrm{~A}][\mathrm{B}]^{2}$

Therefore, the rate of reaction will increase 9 times.

(iii) When the concentrations of both A and B are doubled,

$-\frac{d[\mathrm{R}]}{d t}=k[\mathrm{~A}][\mathrm{B}]^{2}$

$=k[2 \mathrm{~A}][2 \mathrm{~B}]^{2}$

$=8 \cdot k[\mathrm{~A}][\mathrm{B}]^{2}$

Therefore, the rate of reaction will increase 8 times.